整理一下大一上過的離散數(shù)學的實驗代碼...

！已整理至本人的 GitHub上：ljkjk/HNU-DiscreteMathematics-Experiment/湖南大學離散數(shù)學實驗

實驗一

/** 實驗一* 利用字符數(shù)組檢索判斷合式公式的合法性。 * by ljkjk*//* 修改點1：使用萬能頭文件，減少了頭文件的輸入 */ #include <bits/stdc++.h> #define INF 0x3f3f3f3f using namespace std; void rule1(char a[],int i) {if(a[i]>='a' && a[i]<='z') a[i]='1';else if(a[i]=='0') a[i]='1'; } int rule2(char a[],int i) {int n=strlen(a);int _result=0;if(i+1<n && a[i]=='!' && a[i+1]=='1'){a[i]='1';i++;while(a[i+1]!='\0'){a[i]=a[i+1];i++;}a[i]='\0';_result=1;}/* 修改點2：修改運算順序的問題，優(yōu)化了運算的邏輯，減少了出錯的概率 */else if(i+3<n && a[i]=='!' && a[i+1]=='(' && a[i+2]=='1' && a[i+3]==')'){a[i]='1';i++;while(a[i+3]!='\0'){a[i]=a[i+3];i++;}a[i]='\0';_result=1;}else if(i+2<n && a[i]=='(' && a[i+1]=='1' && a[i+2]==')'){a[i]='1';i++;while(a[i+2]!='\0'){a[i]=a[i+2];i++;}a[i]='\0';_result=1;}/* 修改點3：增加遇到空格時的判斷，避免輸入格式問題 */else if(i<n && a[i]==' '){while(a[i+1]!='\0'){a[i]=a[i+1];i++;}a[i]='\0';_result=1;}return _result; } int rule3Con(char a[],int i) {int _result=0;int n=strlen(a);if(i+2<n && a[i]=='1' && a[i+1]=='*' && a[i+2]=='1'){a[i]='1';i++;while(a[i+2]!='\0'){a[i]=a[i+2];i++;}a[i]='\0';_result=1;}return _result; } int rule3BiCond(char a[],int i) {int _result=0;int n=strlen(a);if(i+2<n && a[i]=='1' && a[i+1]=='=' && a[i+2]=='1'){a[i]='1';i++;while(a[i+2]!='\0'){a[i]=a[i+2];i++;}a[i]='\0';_result=1;}return _result; } int rule3Cond(char a[],int i) {int _result=0;int n=strlen(a);if(i+2<n && a[i]=='1' && a[i+1]=='-' && a[i+2]=='1'){a[i]='1';i++;while(a[i+2]!='\0'){a[i]=a[i+2];i++;}a[i]='\0';_result=1;}return _result; } int rule3DisConj(char a[],int i) {int _result=0;int n=strlen(a);if(i+2<n && a[i]=='1' && a[i+1]=='+' && a[i+2]=='1'){a[i]='1';i++;while(a[i+2]!='\0'){a[i]=a[i+2];i++;}a[i]='\0';_result=1;}return _result; } void rule3(char a[],int i) {int n=strlen(a);if(i+2<n && a[i]=='1' && (a[i+1]=='*' || a[i+1]=='-' || a[i+1]=='=' || a[i+1]=='+') && a[i+2]=='1'){a[i]='1';i++;while(a[i+2]!='\0'){a[i]=a[i+2];i++;}a[i]='\0';} }int main() {/* 修改點4：連續(xù)判斷多個公式，方便了使用 */while(true){char pstate[120],pstate0[120];int i=0,nold=0,nnew=0;printf("請輸入公式(析+,合*,條-,雙=,否定!,01):\n");gets(pstate0);fflush(stdin);nold=strlen(pstate0)+1;nnew=strlen(pstate0);for(i=0;i<nnew;i++)pstate[i]=pstate0[i];pstate[i]='\0';i=0;while(i<strlen(pstate)){rule1(pstate,i);i++;}printf("規(guī)則1后：%s\n",pstate);nold=strlen(pstate0)+1;nnew=strlen(pstate);/* 修改點5：優(yōu)化了顯示邏輯，只輸出需要進行的步驟，便于理解 */while(nnew<nold){nold=strlen(pstate);i=0;bool flag=false;while(i<strlen(pstate))if(rule2(pstate,i)==0)i++;else flag=true;if(flag) printf("規(guī)則2后：%s\n",pstate);i=0;flag=false;while(i<strlen(pstate))if(rule3Con(pstate,i)==0)i++;else flag=true;if(flag) printf("規(guī)則3合取后：%s\n",pstate);i=0;flag=false;while(i<strlen(pstate))if(rule3BiCond(pstate,i)==0)i++;else flag=true;if(flag) printf("規(guī)則3雙條件后：%s\n",pstate);i=0;flag=false;while(i<strlen(pstate))if(rule3Cond(pstate,i)==0)i++;else flag=true;if(flag) printf("規(guī)則3單條件后：%s\n",pstate);i=0;flag=false;while(i<strlen(pstate))if(rule3DisConj(pstate,i)==0)i++;else flag=true;if(flag) printf("規(guī)則3析取后：%s\n",pstate);nnew=strlen(pstate);}if(pstate[0]=='1' && strlen(pstate)==1)printf("%s is valid\n\n",pstate0);elseprintf("%s is invalid\n\n",pstate0);}return 0; }

實驗二

/** 實驗二* 輸出合式公式的真值表以及主析取范式* by ljkjk*/#include <bits/stdc++.h> #define INF 0x3f3f3f3f using namespace std; int getalpha(char a[],char b[]) {int n=strlen(a),i=0,j=0,k=0;for(i=0;i<n;++i)if((a[i]>='a'&&a[i]<='z')||(a[i]>='A'&&a[i]<='Z')){/* 修改點1：用strchr函數(shù)替代循環(huán)搜索，優(yōu)化了運行速度 */if(strchr(b,a[i])==NULL){b[j]=a[i];j++;}}/* 修改點2：用sort函數(shù)替代冒泡排序，優(yōu)化了運行速度 */sort(b,b+j);b[j]='\0';return j; }void fillvalue(char a[],char varchar[],int nvar,char valchar[],char resultchar[]) {int nLen=strlen(a),i=0,j=0,k=0;for(i=0;i<nLen;++i)resultchar[i]=a[i];resultchar[i]='\0';for(i=0;i<nLen;++i)for(j=0;j<nvar;++j)if(resultchar[i]==varchar[j]){resultchar[i]=valchar[j];break;} }/* 修改點3：重復(fù)代碼編寫成函數(shù)，減少代碼量 */ void eraser(int r,int &i,int &j,char a[],int w) {if(r==1){j++;while(a[j+w]!='\0'){a[j]=a[j+w];j++;}a[j]='\0';}elsei++; }void negatecal(char a[]) {int _result=0,i=0,j=0;while(i<strlen(a)){j=i;_result=0;if(j+1<strlen(a)&&a[j]=='!'&&a[j+1]=='1'){a[j]='0';_result=1;}else if(j+1<strlen(a)&&a[j]=='!'&&a[j+1]=='0'){a[j]='1';_result=1;}eraser(_result,i,j,a,1);} }void bracket(char a[]) {int _result=0,i=0,j=0;while(i<strlen(a)){j=i;_result=0;if(j+2<strlen(a)&&a[j]=='('&&a[j+1]=='1'&&a[j+2]==')'){a[j]='1';_result=1;}else if(j+2<strlen(a)&&a[j]=='('&&a[j+1]=='0'&&a[j+2]==')'){a[j]='0';_result=1;}eraser(_result,i,j,a,2);} }void con(char a[]) {int _result=0,i=0,j=0;while(i<strlen(a)){j=i;_result=0;if(j+2<strlen(a)&&a[j]=='0'&&a[j+1]=='*'&&a[j+2]=='0'){a[j]='0';_result=1;}else if(j+2<strlen(a)&&a[j]=='0'&&a[j+1]=='*'&&a[j+2]=='1'){a[j]='0';_result=1;}else if(j+2<strlen(a)&&a[j]=='1'&&a[j+1]=='*'&&a[j+2]=='0'){a[j]='0';_result=1;}else if(j+2<strlen(a)&&a[j]=='1'&&a[j+1]=='*'&&a[j+2]=='1'){a[j]='1';_result=1;}eraser(_result,i,j,a,2);} }void bicond(char a[]) {int _result=0,i=0,j=0;while(i<strlen(a)){j=i;_result=0;if(j+2<strlen(a)&&a[j]=='0'&&a[j+1]=='='&&a[j+2]=='0'){a[j]='1';_result=1;}else if(j+2<strlen(a)&&a[j]=='0'&&a[j+1]=='='&&a[j+2]=='1'){a[j]='0';_result=1;}else if(j+2<strlen(a)&&a[j]=='1'&&a[j+1]=='='&&a[j+2]=='0'){a[j]='0';_result=1;}else if(j+2<strlen(a)&&a[j]=='1'&&a[j+1]=='='&&a[j+2]=='1'){a[j]='1';_result=1;}eraser(_result,i,j,a,2);} }void cond(char a[]) {int _result=0,i=0,j=0;while(i<strlen(a)){j=i;_result=0;if(j+2<strlen(a)&&a[j]=='0'&&a[j+1]=='-'&&a[j+2]=='0'){a[j]='1';_result=1;}else if(j+2<strlen(a)&&a[j]=='0'&&a[j+1]=='-'&&a[j+2]=='1'){a[j]='1';_result=1;}else if(j+2<strlen(a)&&a[j]=='1'&&a[j+1]=='-'&&a[j+2]=='0'){a[j]='0';_result=1;}else if(j+2<strlen(a)&&a[j]=='1'&&a[j+1]=='-'&&a[j+2]=='1'){a[j]='1';_result=1;}eraser(_result,i,j,a,2);} }void disconj(char a[]) {int _result=0,i=0,j=0;while(i<strlen(a)){j=i;_result=0;if(j+2<strlen(a)&&a[j]=='0'&&a[j+1]=='+'&&a[j+2]=='0'){a[j]='0';_result=1;}else if(j+2<strlen(a)&&a[j]=='0'&&a[j+1]=='+'&&a[j+2]=='1'){a[j]='1';_result=1;}else if(j+2<strlen(a)&&a[j]=='1'&&a[j+1]=='+'&&a[j+2]=='0'){a[j]='1';_result=1;}else if(j+2<strlen(a)&&a[j]=='1'&&a[j+1]=='+'&&a[j+2]=='1'){a[j]='1';_result=1;}eraser(_result,i,j,a,2);} }/* 修改點4：標準化輸入，避免輸入格式錯誤 */ void getformula(char a[]) {char cc;int cnt=0;while((cc=getchar())!='\n')if((cc>='a'&&cc<='z')||(cc>='A'&&cc<='Z')||cc=='='||cc=='-'||cc=='+'||cc=='*'||cc=='!'||cc=='0'||cc=='1'||cc=='('||cc==')')a[cnt++]=cc; }int main() {/* 修改點5：持續(xù)化輸入，方便使用 */while(1){char pstate[120]={' '},pstate0[120]={' '},charlist[120]={' '},charval[120]={' '};char minitem[1024][52]={' '},maxitem[1024][52]={' '},truetable[1024]={' '};int i=0,nold=0,nnew=0,nvar=1,nrow=1,j=0,flagsum=1,iminitem=0,imaxitem=0;printf("請輸入公式(析+,合*,條-,雙=,否定!,01):\n");//gets(pstate0);getformula(pstate0);fflush(stdin);nold=strlen(pstate0)+1;nnew=strlen(pstate0);for(i=0;i<nnew;++i)pstate[i]=pstate0[i];pstate[i]='\0';nvar=getalpha(pstate,charlist);nrow=1;for(i=0;i<nvar;++i){charval[i]='0';/* 修改點6：用位運算加速乘法，優(yōu)化了運行速度 */nrow<<=1;}charval[i]='\0';printf("\n");for(i=0;i<nvar;++i)printf("%4c",charlist[i]);printf("%15c%s\n",' ',pstate);for(i=0;i<nvar;++i)printf("%4c",'-');printf("|");for(i=0;i<60;++i)printf("%c",'-');printf("\n");int sum=1;for(i=0;i<nrow;++i){for(j=0;j<nvar;++j)printf("%4c",charval[j]);pstate[0]='\0';fillvalue(pstate0,charlist,nvar,charval,pstate);nold=strlen(pstate0)+1;nnew=strlen(pstate);while(nnew<nold){nold=strlen(pstate);negatecal(pstate);bracket(pstate);con(pstate);bicond(pstate);cond(pstate);disconj(pstate);nnew=strlen(pstate);}if(strlen(pstate)==1){if(pstate[0]=='1'){for(j=0;j<nvar;++j)minitem[iminitem][j]=charval[j];minitem[iminitem][j]='\0';iminitem++;}else if(pstate[0]=='0'){for(j=0;j<nvar;++j)maxitem[imaxitem][j]=charval[j];maxitem[imaxitem][j]='\0';imaxitem++;}truetable[i]=pstate[0];}printf("%20c%s",' ',pstate);printf("\n");/* 修改點7：優(yōu)化為變元賦值的邏輯，優(yōu)化了運行速度 */int tmp=sum;for(int j=nvar-1;j>=0;--j){charval[j]=(tmp%2+'0');tmp/=2;}sum++;}printf("\n");printf("主析取范式：\n");for(i=0;i<iminitem;++i)if(i==0)printf("m%s",minitem[i]);elseprintf("+m%s",minitem[i]);printf("\n");for(i=0;i<iminitem;++i)if(i==0){printf("(");for(j=0;j<nvar;++j){if(j==0)if(minitem[i][j]=='1')printf("%c",charlist[j]);elseprintf("!%c",charlist[j]);elseif(minitem[i][j]=='1')printf("*%c",charlist[j]);elseprintf("*!%c",charlist[j]);}printf(")");}else{printf("+(");for(j=0;j<nvar;++j){if(j==0)if(minitem[i][j]=='1')printf("%c",charlist[j]);elseprintf("!%c",charlist[j]);elseif(minitem[i][j]=='1')printf("*%c",charlist[j]);elseprintf("*!%c",charlist[j]);}printf(")");}printf("\n\n");/* 修改點8：添加主合取范式的輸出，豐富了功能 */printf("主合取范式：\n");for(i=0;i<imaxitem;++i)if(i==0)printf("M%s",maxitem[i]);elseprintf("*M%s",maxitem[i]);printf("\n");for(i=0;i<imaxitem;++i)if(i==0){printf("(");for(j=0;j<nvar;++j){if(j==0)if(maxitem[i][j]=='0')printf("%c",charlist[j]);elseprintf("!%c",charlist[j]);elseif(maxitem[i][j]=='0')printf("+%c",charlist[j]);elseprintf("+!%c",charlist[j]);}printf(")");}else{printf("*(");for(j=0;j<nvar;++j){if(j==0)if(maxitem[i][j]=='0')printf("%c",charlist[j]);elseprintf("!%c",charlist[j]);elseif(maxitem[i][j]=='0')printf("+%c",charlist[j]);elseprintf("+!%c",charlist[j]);}printf(")");}printf("\n\n");}return 0; }

實驗三

/** 實驗三* 利用程序進行自然推理* by ljkjk*//* 修改點1：萬能頭文件，減少了頭文件的輸入 */ #include <bits/stdc++.h> using namespace std; struct tmd{char gs[120],gslast[120],reason[120];int nletter,nused,isletter,iscond; };void nonoop2(char aa[]) {int i=0,j=0;while(i<strlen(aa)-2){if(i+1<strlen(aa)&&aa[i]=='!'&&aa[i+1]=='!'){j=i;while(j<strlen(aa)-2){aa[j]=aa[j+2];j++;}aa[j]='\0';break;}elsei++;} }void print(tmd tmrec[],int np) {int i=0;for(i=0;i<np;++i){if(tmrec[i].isletter==1)printf("(%d)\t%s為真\t\t\t%s---文字\n",i+1,tmrec[i].gs,tmrec[i].reason);else if(tmrec[i].iscond==1)printf("(%d)\t%s-%s為真\t\t\t%s---條件式\n",i+1,tmrec[i].gs,tmrec[i].gslast,tmrec[i].reason);elseprintf("(%d)\t%s為真\t\t\t%s\n",i+1,tmrec[i].gs,tmrec[i].reason);} }int setreason(tmd tmrec[],int np,string r0,int j0,int j1,int nused0,int iscond0,int isletter0) {char stmpj0[20],stmpj1[20];int nlen1=0,j=0,nlenj0=0,nlenj1=0;nlen1=r0.length();itoa(j0+1,stmpj0,10);nlenj0=strlen(stmpj0);itoa(j1+1,stmpj1,10);nlenj1=strlen(stmpj1);if(j0==-1)//輸出“前提條件”{for(j=0;j<nlen1;++j)tmrec[np].reason[j]=r0[j];tmrec[np].reason[j]='\0';}else if(j1==-1){tmrec[np].reason[0]='(';for(j=0;j<nlenj0;++j)tmrec[np].reason[j+1]=stmpj0[j];tmrec[np].reason[j+1]=')';for(j=0;j<nlen1;++j)tmrec[np].reason[j+2+nlenj0]=r0[j];tmrec[np].reason[j+2+nlenj0]='\0';}else{tmrec[np].reason[0]='(';for(j=0;j<nlenj0;++j)tmrec[np].reason[j+1]=stmpj0[j];tmrec[np].reason[j+1]=')';tmrec[np].reason[nlenj0+2]='(';for(j=0;j<nlenj1;++j)tmrec[np].reason[nlenj0+3+j]=stmpj1[j];tmrec[np].reason[nlenj0+3+j]=')';for(j=0;j<nlen1;++j)tmrec[np].reason[nlenj0+nlenj1+4+j]=r0[j];tmrec[np].reason[nlenj0+nlenj1+4+j]='\0';}tmrec[np].nused=nused0;tmrec[np].iscond=iscond0;tmrec[np].isletter=isletter0; }/* 修改點2：調(diào)換了inputprimary函數(shù)與setreason函數(shù)的位置，解決了報錯的問題 */ int inputprimary(tmd gs0[]) {tmd tmp;char pstate[120];/* 修改點3：用string替換字符數(shù)組，消除了警告 */string r0="前提條件";string r1="原命題的逆否命題";string r2="雙條件導出的條件式";string r3="析取式轉(zhuǎn)換為條件式";int i=0,j=0,nlen=0,k=0;int i0=0;printf("輸完一個前提條件請按回車，不輸直接回車則結(jié)束\n析+,合-,條-,雙=,否定!\n前提中只能為雙條件、單條件、析取式，\n若為兩個個條件的合取，請輸入兩個前提，文字請在最前面輸入：\n");while(1){gets(pstate);nlen=strlen(pstate);if(nlen==0)break;if(nlen==1){gs0[i].nletter=strlen(pstate);gs0[i].gs[0]=pstate[0];gs0[i].gs[1]='\0';gs0[i].gslast[0]='\0';setreason(gs0,i,r0,-1,-1,0,0,1);}else if(nlen==2&&pstate[0]=='!'){gs0[i].nletter=strlen(pstate);gs0[i].gs[0]=pstate[0];gs0[i].gs[1]=pstate[1];gs0[i].gs[2]='\0';gs0[i].gslast[0]='\0';setreason(gs0,i,r0,-1,-1,0,0,1);}else{for(j=0;j<nlen;++j){if(pstate[j]=='-'){gs0[i].nletter=strlen(pstate);for(k=0;k<j;++k)gs0[i].gs[k]=pstate[k];gs0[i].gs[k]='\0';for(k=j+1;k<nlen;++k)gs0[i].gslast[k-j-1]=pstate[k];gs0[i].gslast[k-j-1]='\0';setreason(gs0,i,r0,-1,-1,0,1,0);i++;gs0[i].nletter=gs0[i-1].nletter;gs0[i].gslast[0]='!';for(k=0;k<j;++k)gs0[i].gslast[k+1]=pstate[k];gs0[i].gslast[k+1]='\0';nonoop2(gs0[i].gslast);gs0[i].gs[0]='!';for(k=j+1;k<nlen;++k)gs0[i].gs[k-j-1+1]=pstate[k];//>??????gs0[i].gs[k-j-1+1]='\0';nonoop2(gs0[i].gs);setreason(gs0,i,r1,i-1,-1,0,1,0);break;}else if(pstate[j]=='='){gs0[i].nletter=strlen(pstate);for(k=0;k<strlen(pstate);++k)gs0[i].gs[k]=pstate[k];gs0[i].gs[k]='\0';gs0[i].gslast[0]='\0';setreason(gs0,i,r0,-1,-1,0,0,0);i++;gs0[i].nletter=strlen(pstate);for(k=0;k<j;++k)gs0[i].gslast[k-j-1]=pstate[k];gs0[i].gslast[k-j-1]='\0';setreason(gs0,i,r2,i-1,-1,0,1,0);i++;gs0[i].nletter=gs0[i-1].nletter;for(k=0;k<j;++k)gs0[i].gslast[k]=pstate[k];gs0[i].gs[k-j-1]='\0';setreason(gs0,i,r2,i-2,-1,0,1,0);break;}else if(pstate[j]=='+'){gs0[i].nletter=strlen(pstate);for(k=0;k<strlen(pstate);++k)gs0[i].gs[k]=pstate[k];gs0[i].gs[k]='\0';gs0[i].gslast[0]='\0';setreason(gs0,i,r0,-1,-1,0,0,0);i++;gs0[i].nletter=strlen(pstate);gs0[i].gs[0]='!';for(k=0;k<j;++k)gs0[i].gs[k+1]=pstate[k];gs0[i].gs[k+1]='\0';for(k=j+1;k<nlen;++k)gs0[i].gslast[k-j-1]=pstate[k];gs0[i].gslast[k-j-1]='\0';setreason(gs0,i,r3,i-1,-1,0,1,0);nonoop2(gs0[i].gs);i++;gs0[i].nletter=strlen(pstate);for(k=0;k<j;++k)gs0[i].gslast[k]=pstate[k];gs0[i].gslast[k]='\0';gs0[i].gs[0]='!';for(k=j+1;k<nlen;++k)gs0[i].gs[k-j-1+1]=pstate[k];gs0[i].gs[k-j-1+1]='\0';setreason(gs0,i,r3,i-2,-1,0,1,0);nonoop2(gs0[i].gs);break;}}if(j>=nlen){gs0[i].nletter=strlen(pstate);for(k=0;k<nlen;++k)gs0[i].gs[k]=pstate[k];gs0[i].gs[k]='\0';gs0[i].gslast[0]='\0';setreason(gs0,i,r0,-1,-1,0,0,0);}}i++;}nlen=i;for(i=0;i<nlen-1;++i){k=i;/* 修改點4：將nlen-1改為nlen，否則最后一個元素無法排序 */for(j=i+1;j<nlen;++j)if(gs0[k].nletter>gs0[j].nletter)k=j;/* 修改點5：用swap替代中間元交換兩個數(shù)的值，使代碼簡潔 */if(k>i) swap(gs0[i],gs0[k]);}return nlen; }int main() {/* 修改點6：連續(xù)輸入 */while(1){tmd gs0[100];char result0[128];tmd tmrec[1024];char stmp[128];char lastreason[128]="";string r01="假言推理";int i=0,j=0,k=0;int np=1,np0=0,is0k=0;int i0=0,nposletter=0,nposcond=0;np0=inputprimary(gs0);printf("輸入要推理的結(jié)論，結(jié)論只能是文字，\n若是條件式、析取式請先手工轉(zhuǎn)換為條件，將前件作為附加前提：\n");gets(result0);fflush(stdin);for(i=0;i<np0;++i)tmrec[i]=gs0[i];np=i;nposletter=0;nposcond=0;is0k=0;i0=-1;while(1){i=i0+1;while(i<np&&tmrec[i].isletter!=1)i++;if(i>=np)break;i0=i;nposletter=i;strcpy(stmp,tmrec[i].gs);np0=np-1;while(np>np0){np0=np;for(i=0;i<np;++i)if(tmrec[i].iscond==1&&tmrec[i].nused==0)break;if(i==np)break;while(i<np){if(tmrec[i].iscond==1)if((strcmp(lastreason,tmrec[i].reason)!=0)||(strcmp(lastreason,tmrec[i].reason)==0&&tmrec[i].reason[0]!='('))if(strcmp(tmrec[i].gs,stmp)==0){strcpy(lastreason,tmrec[i].reason);tmrec[nposletter].nused++;tmrec[i].nused++;strcpy(stmp,tmrec[i].gslast);strcpy(tmrec[np].gs,stmp);tmrec[np].gslast[0]='\0';setreason(tmrec,np,r01,nposletter,i,0,0,1);nposletter=np;np++;if(strcmp(result0,stmp)==0){is0k=1;break;}}i++;}}if(is0k==1)break;}if(is0k==1)printf("推理成立，推理過程如下：\n");elseprintf("推理不成立，推理過程如下:\n");print(tmrec,np);printf("\n\n");}return 0; }

實驗四

/** 實驗四* 利用消解法進行推理* by ljkjk*/#include <string.h> #include <iostream> #include <cstdio> #include <cmath> #include <cstdlib> #include <algorithm>struct tmd {char gs[120], gsLast[120], niyou[120];//前件與后件及理由int nLitter, nUsed, isLitter, isCond; };void nonoop2( char aa[] ) {//!!p=pint i = 0, j = 0;while (i < strlen( aa ) - 2){//至少還有兩個字符if (((i + 1) < strlen( aa )) && (aa[i] == '!') && (aa[i + 1] == '!')){j = i;while (j < strlen( aa ) - 2){aa[j] = aa[j + 2];j++;}aa[j] = '\0';break;}elsei++;} }void printYsh( struct tmd tmdrec[], int np ) {int i = 0;for (i = 0; i < np; i++){if (tmdrec[i].isLitter == 1)printf( "(%d)\t%s為真\t\t\t%s---文字\n", i + 1, tmdrec[i].gs, tmdrec[i].niyou );else if (tmdrec[i].isCond == 1)printf( "(%d)\t%s+%s為真\t\t\t%s---析取式\n", i + 1, tmdrec[i].gs, tmdrec[i].gsLast, tmdrec[i].niyou );elseprintf( "(%d)\t%s為真\t\t\t%s\n", i + 1, tmdrec[i].gs, tmdrec[i].niyou );} }/* 修改點1：調(diào)換函數(shù)位置，否則會出現(xiàn)調(diào)用未定義函數(shù)的情況 */ int setNiyou( struct tmd tmdrec[], int np, char ny0[], int j0, int j1, int nUsed0, int isCond0, int isLitter0 ) {//將字符串ny0與j0賦到推理意見中char stmdpj0[20], stmdpj1[20];int nLen1 = 0, j = 0, nLenj0 = 0, nLenj1 = 0;nLen1 = strlen( ny0 );itoa( j0 + 1, stmdpj0, 10 );nLenj0 = strlen( stmdpj0 );//前一個依據(jù)itoa( j1 + 1, stmdpj1, 10 );nLenj1 = strlen( stmdpj1 );//后一個依據(jù)if (j0 == -1){//原始前提for (j = 0; j < nLen1; j++)tmdrec[np].niyou[j] = ny0[j];tmdrec[np].niyou[j] = '\0';}else if (j1 == -1)//由前一步推理所得結(jié)論{tmdrec[np].niyou[0] = '(';for (j = 0; j < nLenj0; j++)tmdrec[np].niyou[j + 1] = stmdpj0[j];tmdrec[np].niyou[j + 1] = ')';for (j = 0; j < nLen1; j++)tmdrec[np].niyou[j + 2 + nLenj0] = ny0[j];tmdrec[np].niyou[j + 2 + nLenj0] = '\0';}else{//由前二步推理所得tmdrec[np].niyou[0] = '(';for (j = 0; j < nLenj0; j++)tmdrec[np].niyou[j + 1] = stmdpj0[j];tmdrec[np].niyou[j + 1] = ')';tmdrec[np].niyou[nLenj0 + 2] = '(';for (j = 0; j < nLenj1; j++)tmdrec[np].niyou[nLenj0 + 3 + j] = stmdpj1[j];tmdrec[np].niyou[nLenj0 + 3 + j] = ')';for (j = 0; j < nLen1; j++)tmdrec[np].niyou[nLenj0 + nLenj1 + 4 + j] = ny0[j];tmdrec[np].niyou[nLenj0 + nLenj1 + 4 + j] = '\0';}tmdrec[np].nUsed = nUsed0;//附加前提從未使用過nUsed0,int isCond0,int isLitter0tmdrec[np].isCond = isCond0;//是條件式tmdrec[np].isLitter = isLitter0;//是文字 }void swaptmd(tmd &a,tmd &b) {tmd temp;temp=a;a=b;b=temp; }int inputPrimary( struct tmd gs0[] ) {struct tmd tmdp;char pstate[120];char *ny0 = "前提條件";char *ny1 = "條件式轉(zhuǎn)為析取式";char *ny2 = "雙條件導出的析取式";int i = 0, j = 0, nLen = 0, k = 0;int i0 = 0;//原始條件printf( "輸完一個前提條件請按回車，不輸直接回車則結(jié)束\n析+,合*,條-,雙=,否定!\n前提中只能為雙條件、單條件、析取式,\n若為2個條件的合取,請輸入2個前提,文字請在最前面輸入:\n" );while (1){gets( pstate );nLen = strlen( pstate );if (nLen == 0){break;}//設(shè)置nUsed,isLitter,isCond,nLittle的值//判斷是否為文字if (nLen == 1)//標注單個文字{gs0[i].nLitter = strlen( pstate );gs0[i].gs[0] = pstate[0];gs0[i].gs[1] = '\0';gs0[i].gsLast[0] = '\0';setNiyou( gs0, i, ny0, -1, -1, 0, 0, 1 );//前提類型，無，無，未使用，不是條件式，是文字}else if ((nLen == 2) && (pstate[0] == '!')) //標注！p{gs0[i].nLitter = strlen( pstate );gs0[i].gs[0] = pstate[0];gs0[i].gs[1] = pstate[1];gs0[i].gs[2] = '\0';gs0[i].gsLast[0] = '\0';setNiyou( gs0, i, ny0, -1, -1, 0, 0, 1 );//前提類型，無，無，未使用，不是條件式，是文字}else{for (j = 0; j < nLen; j++){if (pstate[j] == '-')//標注條件式p-q{gs0[i].nLitter = strlen( pstate );for (k = 0; k < nLen; k++)gs0[i].gs[k] = pstate[k];//整個表達式進入gsgs0[i].gs[k] = '\0';gs0[i].gsLast[0] = '\0';setNiyou( gs0, i, ny0, -1, -1, 0, 0, 0 );//前提類型，無，無，未使用，不是析取式，不是文字i++;gs0[i].nLitter = gs0[i - 1].nLitter;gs0[i].gs[0] = '!';for (k = 0; k < j; k++)gs0[i].gs[k + 1] = pstate[k];gs0[i].gs[k + 1] = '\0';nonoop2( gs0[i].gs );for (k = j + 1; k < nLen; k++)gs0[i].gsLast[k - j - 1] = pstate[k];gs0[i].gsLast[k - j - 1] = '\0';setNiyou( gs0, i, ny1, i - 1, -1, 0, 1, 0 );//前提類型，無，無，未使用，是條件式，不是文字break;}else if (pstate[j] == '=')//標注雙條件p=q{//先保存雙條件gs0[i].nLitter = strlen( pstate );for (k = 0; k < strlen( pstate ); k++) { gs0[i].gs[k] = pstate[k]; }//雙條件全部進gsgs0[i].gs[k] = '\0';gs0[i].gsLast[0] = '\0';setNiyou( gs0, i, ny0, -1, -1, 0, 0, 0 );//前提類型，無，無，未使用，是條件式，不是文字//p-q即!p+qi++;gs0[i].nLitter = strlen( pstate );gs0[i].gs[0] = '!';for (k = 0; k < j; k++) { gs0[i].gs[k + 1] = pstate[k]; }//p進gsgs0[i].gs[k + 1] = '\0';for (k = j + 1; k < nLen; k++) { gs0[i].gsLast[k - j - 1] = pstate[k]; }//q進gsLastgs0[i].gsLast[k - j - 1] = '\0';setNiyou( gs0, i, ny2, i - 1, -1, 0, 1, 0 );//前提類型，無，無，未使用，是條件式，不是文字nonoop2( gs0[i].gs );//去掉可能存在的!!?//q-p=p+!qi++;gs0[i].nLitter = gs0[i - 1].nLitter;for (k = 0; k < j; k++) { gs0[i].gs[k] = pstate[k]; }//條件前件p進gsgs0[i].gs[k] = '\0';gs0[i].gsLast[0] = '!';for (k = j + 1; k < nLen; k++) { gs0[i].gsLast[k - j - 1 + 1] = pstate[k]; }//條件后件!q進gsLastgs0[i].gsLast[k - j - 1 + 1] = '\0';setNiyou( gs0, i, ny2, i - 2, -1, 0, 1, 0 );//前提類型，無，無，未使用，是條件式，不是文字nonoop2( gs0[i].gsLast );//去掉可能存在的!!?break;}else if (pstate[j] == '+')//標注析取式p+q，也要分解到gs與gsLast中{gs0[i].nLitter = strlen( pstate );for (k = 0; k < j; k++)gs0[i].gs[k] = pstate[k]; //前件進gsgs0[i].gs[k] = '\0';for (k = j + 1; k < nLen; k++)gs0[i].gsLast[k - j - 1] = pstate[k]; //條件全部進gsgs0[i].gsLast[k - j - 1] = '\0';setNiyou( gs0, i, ny0, -1, -1, 0, 1, 0 );//前提類型,無，無，未使用，是條件式，不是文字break;}}if (j >= nLen)//不是條件式，也不是文字，則是普通的條件{gs0[i].nLitter = strlen( pstate );for (k = 0; k < nLen; k++)gs0[i].gs[k] = pstate[k]; //公式全進gsgs0[i].gs[k] = '\0';gs0[i].gsLast[0] = '\0'; //gsLast為空串setNiyou( gs0, i, ny0, -1, -1, 0, 0, 0 );//前提類型，無，無，未使用，不是條件式，不是文字}}i++;//當前公式處理完以后，指針i的值增1}nLen = i;//按字符串的長度排序for (i = 0; i < nLen - 1; i++){k = i;//for (j = i + 1; j < nLen - 1; j++)/* 修改點2：修改循環(huán)條件使得最后的元素可排序 */for (j = i + 1; j < nLen; j++)if (gs0[k].nLitter > gs0[j].nLitter)k = j;if (k > i){/* 修改點3：使用自定義的swaptmd排序 */swaptmd(gs0[i],gs0[k]); // tmdp = gs0[i]; // gs0[i] = gs0[k]; // gs0[k] = tmdp;}}return nLen; }int main() {struct tmd gs0[100];//推理前提條件char result0[128]; //結(jié)論struct tmd tmdrec[1024];//最多1000步char stmdp[128];char lastNiYou[128] = " ";//上一個推理式的理由char *ny01 = "消解";int i = 0, j = 0, k = 0;int np = 1, np0 = 0, isOk = 0;int i0 = 0, nPosLitter = 0, nPosCond = 0;//文字起始的位置，首個文字的位置，消解式的位置np0 = inputPrimary( gs0 );//輸入結(jié)論printf( "輸入要推理的結(jié)論，結(jié)論只能是文字，\n若是條件式，析取式請先手工轉(zhuǎn)換為條件，將前件作為附加前提：" );gets( result0 );fflush( stdin );for (i = 0; i < np0; i++){tmdrec[i] = gs0[i];//所有原始公式轉(zhuǎn)抄到tmdrec中}np = i;//推理隊列的尾部指針nPosLitter = 0;//文字的位置號nPosCond = 0;//條件的位置號isOk = 0;i0 = -1;while (1){i = i0 + 1;//尋找下一個文字，i是起始位置，np是命令串的長度while ((i < np) && (tmdrec[i].isLitter != 1))i++;if (i >= np)break;//找不到文字我就沒法推理了i0 = i;//記錄從源頭查詢的首個文字的位置號，下次從此號往后尋找nPosLitter = i;//記錄文字的位置strcpy( stmdp, tmdrec[i].gs );//保存當前文字的內(nèi)容np0 = np - 1;while (np > np0) //從當前文字的下一個位置起尋找析取式，則一路往下走{np0 = np;for (i = 0; i < np; i++)//找到一個沒有用過的戲曲式if ((tmdrec[i].isCond == 1) && (tmdrec[i].nUsed == 0))break;if (i == np)break;//沒有找到則結(jié)束推理，所有條件式都用到了while (i < np)//若找到了這樣的條件式{if ((tmdrec[i].isCond == 1))//若是條件式{//與上條命令的來源不同，或者但是同為前提條件也是可以的，即首個字符不是（if (((strcmp( lastNiYou, tmdrec[i].niyou ) != 0) || ((strcmp( lastNiYou, tmdrec[i].niyou ) == 0) && tmdrec[i].niyou[0] != '('))){if ((tmdrec[i].gs[0] == '!') && (stmdp[0] != '!') && (strlen( tmdrec[i].gs ) - strlen( stmdp ) == 1)) // !p+q p cuo{//如果析取式的前件與stmdp即可消解，則將后件保存的stmdp中j = 0;while (j < strlen( stmdp ))//依次比較每個字符{if (tmdrec[i].gs[j + 1] != stmdp[j])break;//有一個不相等則結(jié)束比較j++;}if (j >= strlen( stmdp )) //如果比到最后仍然相等，則這二個可消解{strcpy( lastNiYou, tmdrec[i].niyou );tmdrec[nPosLitter].nUsed++; //這個文字用過一次了tmdrec[i].nUsed++; //這個析取式用過一次了strcpy( stmdp, tmdrec[i].gsLast ); //將次消解結(jié)果保存到推導序列中strcpy( tmdrec[np].gs, stmdp ); //將當前推出來的結(jié)果保存起來tmdrec[np].gsLast[0] = '\0'; //后件清空，保存當前條件setNiyou( tmdrec, np, ny01, nPosLitter, i, 0, 0, 1 ); //前提類型，有，無，未使用，不是條件nPosLitter = np; //記錄當前文字的序號np++;if (strcmp( result0, stmdp ) == 0){isOk = 1; //推出結(jié)論同條原是調(diào)節(jié)的下一輪break;}}}else if ((tmdrec[i].gsLast[0] == '!') && (stmdp[0] != '!') && (strlen( tmdrec[i].gsLast ) - strlen( stmdp ) == 1)) //a+!b b dui{//如果析取式的后件與stmdp即可消解，則將前件保存到stmdp中j = 0;while (j < strlen( stmdp )) //依次比較每一個字符{if (tmdrec[i].gsLast[j + 1] != stmdp[j])break; //有一個不相等則結(jié)束比較j++;}if (j >= strlen( stmdp )) //如果比到最后仍然相等，則這兩個可消解{strcpy( lastNiYou, tmdrec[i].niyou );tmdrec[nPosLitter].nUsed++; //這個文字用過一次了tmdrec[i].nUsed++; //這個析取式用過一次了strcpy( stmdp, tmdrec[i].gs ); //將次消解結(jié)果保存到推導序列中strcpy( tmdrec[np].gs, stmdp ); //將當前推出來的結(jié)果保存起來tmdrec[np].gsLast[0] = '\0'; //后件清空，保存當前條件setNiyou( tmdrec, np, ny01, nPosLitter, i, 0, 0, 1 ); //前提類型，有，無，未使用，不是條件nPosLitter = np; //記錄當前文字的序號np++;if (strcmp( result0, stmdp ) == 0){isOk = 1; //推出結(jié)論同條原是調(diào)節(jié)的下一輪break;}}}else if ((tmdrec[i].gs[0] != '!') && (stmdp[0] == '!') && (strlen( tmdrec[i].gs ) - strlen( stmdp ) == -1)) // p+q !p{//如果析取式的后件與stmdp即可消解，則將前件保存到stmdp中j = 0;while (j < strlen( tmdrec[i].gs )) //依次比較每一個字符{if (stmdp[j + 1] != tmdrec[i].gs[j])break; //有一個不相等則結(jié)束比較j++;}if (j >= strlen( tmdrec[i].gs )){strcpy( lastNiYou, tmdrec[i].niyou );tmdrec[nPosLitter].nUsed++; //這個文字用過一次了tmdrec[i].nUsed++; //這個析取式用過一次了strcpy( stmdp, tmdrec[i].gsLast ); //將次消解結(jié)果保存到推導序列中strcpy( tmdrec[np].gs, stmdp ); //將當前推出來的結(jié)果保存起來tmdrec[np].gsLast[0] = '\0'; //后件清空，保存當前條件setNiyou( tmdrec, np, ny01, nPosLitter, i, 0, 0, 1 ); //前提類型，有，無，未使用，不是條件nPosLitter = np; //記錄當前文字的序號np++;if (strcmp( result0, stmdp ) == 0){isOk = 1; //推出結(jié)論同條原是調(diào)節(jié)的下一輪break;}}}else if ((tmdrec[i].gsLast[0] != '!') && (stmdp[0] == '!') && (strlen( tmdrec[i].gsLast ) - strlen( stmdp ) == -1)) //p+q !q{//如果析取式的后件與stmdp即可消解，則將前件保存到stmdp中j = 0;while (j < strlen( tmdrec[i].gsLast ))//依次比較每一個字符{if (stmdp[j + 1] != tmdrec[i].gsLast[j])break; //有一個不相等則結(jié)束比較j++;}if (j >= strlen( tmdrec[i].gsLast ))//如果比到最后仍然相等，則這兩個可消解{strcpy( lastNiYou, tmdrec[i].niyou );tmdrec[nPosLitter].nUsed++; //這個文字用過一次了tmdrec[i].nUsed++; //這個條件用過一次了strcpy( stmdp, tmdrec[i].gs ); //將此中間結(jié)果保存到推導序列中strcpy( tmdrec[np].gs, stmdp ); //將當前推出來的結(jié)果保存起來tmdrec[np].gsLast[0] = '\0'; //后件清空，保存當前條件setNiyou( tmdrec, np, ny01, nPosLitter, i, 0, 0, 1 );//前提類型，有，無，未使用，不是條件式nPosLitter = np; //記錄當前文字的序號np++;if (strcmp( result0, stmdp ) == 0){isOk = 1; //推出結(jié)論同原始條件的下一輪break;}}}}}i++;//判斷下一個表達式是否為條件，是否為可推理的條件式}}if (isOk == 1)break; //我推出來了，不要再找下一個文字了}if (isOk == 1)printf( "我推出來了,推理過程如下:\n" );elseprintf( "我推不出來,推理過程如下:\n" );printYsh( tmdrec, np ); }

實驗五?

/** 實驗五* 計算兩個集合的若干運算* by ljkjk*/#include <iostream> #include <cmath> #include <ctime> #include <cstdio> #include <cstdlib> #include <string> #include <string.h> #include <vector> #include <set> #include <map> #include <stack> #include <queue> #include <iomanip> #include <algorithm> #define INF 0x3f3f3f3f using namespace std; const int max0 = 1e8; /* 修改點1：改進print函數(shù)，優(yōu)化了代碼質(zhì)量 */ int print(char a[]) { // int nlen=0; // int i=0; // nlen=strlen(a); // if(nlen>0) // printf("%c",a[0]); // for(i=1;i<nlen;++i) // printf(",%c",a[i]);int nlen=strlen(a);if(nlen==0)printf("空集");elsefor(int i=0;i<nlen;++i){if(i>0) printf(",");printf("%c",a[i]);}printf("\n");return nlen; }int input(char a[]) {char nlen=0,i=0,j=0,k=0;char stmp[1024];printf("集合元素只能是A-Z,a-z,0-9,其他字符被當作分隔符去掉：\n");gets(stmp);fflush(stdin);nlen=strlen(stmp);for(i=0;i<nlen;++i)if((stmp[i]>='A'&&stmp[i]<='Z')||(stmp[i]>='a'&&stmp[i]<='z')||(stmp[i]>='0'&&stmp[i]<='9'))/* 修改點2：使用strchr函數(shù)代替循環(huán)搜索，優(yōu)化了運行速度 */if(strchr(a,stmp[i])==NULL){a[j]=stmp[i];j++;}/* 修改點3：使用sort函數(shù)將集合排序，方便了使用 */sort(a,a+j);a[j]='\0';return j; }int setjiao(char a[],char b[],char c[]) {int i=0,j=0,k=0,nlen1=0,nlen2=0;nlen1=strlen(a);nlen2=strlen(b);for(i=0;i<nlen1;++i)if(strchr(b,a[i])!=NULL){c[k]=a[i];++k;}sort(c,c+k);c[k]='\0';return k; }int setbin(char a[],char b[],char c[]) {int i=0,j=0,k=0,nlen1=0,nlen2=0;nlen1=strlen(a);nlen2=strlen(b);/* 修改點4：使用strcpy函數(shù)復(fù)制數(shù)組 */strcpy(c,a);k=nlen1;for(i=0;i<nlen2;++i)if(strchr(c,b[i])==NULL){c[k]=b[i];++k;}sort(c,c+k);c[k]='\0';return k; }int setsub(char a[],char b[],char c[]) {int i=0,j=0,k=0,nlen1=0,nlen2=0;nlen1=strlen(a);nlen2=strlen(b);for(i=0;i<nlen1;++i)if(strchr(b,a[i])==NULL){c[k]=a[i];++k;}sort(c,c+k);c[k]='\0';return k; }int setdc(char a[],char b[],char c[]) {int i=0,j=0,k=0,nlen1=0,nlen2=0;nlen1=strlen(a);nlen2=strlen(b);for(i=0;i<nlen1;++i){for(j=0;j<nlen2;++j)if(a[i]==b[j])break;if(j>=nlen1){c[k]=a[i];++k;}}for(i=0;i<nlen2;++i){for(j=0;j<nlen1;++j)if(b[i]==a[j])break;if(j>=nlen1){c[k]=b[i];++k;}}c[k]='\0';return k; }int setdicarl(char a[],char b[],char c[][3]) {int i=0,j=0,k=0,nlen1=0,nlen2=0;nlen1=strlen(a);nlen2=strlen(b);for(i=0;i<nlen1;++i){for(j=0;j<nlen2;++j){c[k][0]=a[i];c[k][1]=b[j];c[k][2]='\0';++k;}}return k; }int main() {char a[1024],b[1024],c[1024],dcar[1024][3];int nlen=0;int i=0;printf("輸入集合a：");input(a);printf("輸入集合b：");input(b);printf("A:");print(a);printf("B:");print(b);printf("集合的交:");setjiao(a,b,c);print(c);printf("集合的并:");setbin(a,b,c);print(c);printf("集合的差:");setsub(a,b,c);print(c);printf("集合的對稱差:");setdc(a,b,c);print(c);nlen=setdicarl(a,b,dcar);if(nlen>0)printf("<%c,%c>",dcar[0][0],dcar[0][1]);for(i=1;i<nlen;++i)printf(",<%c,%c>",dcar[i][0],dcar[i][1]);return 0; }

實驗六

#include <iostream> #include <cmath> #include <ctime> #include <cstdio> #include <cstdlib> #include <string> #include <string.h> #include <vector> #include <set> #include <map> #include <stack> #include <queue> #include <iomanip> #include <algorithm> #define INF 0x3f3f3f3f #define N 100 using namespace std; struct stree{int pointa,pointb,weight; };/* 修改點1：合并兩個矩陣的輸入 */ int inputadjaceedge(int a[][N],int b[][N],int *n,int *m) {int i=0,j=0;printf("點數(shù)：");scanf("%d",n);fflush(stdin);printf("邊數(shù)：");scanf("%d",m);fflush(stdin);printf("n=%d m=%d\n",*n,*m);printf("\n輸入鄰接矩陣的值：");for(i=0;i<*n;++i)for(j=0;j<*n;++j)scanf("%d",&a[i][j]);fflush(stdin);printf("\n輸入關(guān)聯(lián)矩陣的值：");for(i=0;i<*n;++i)for(j=0;j<*m;++j)scanf("%d",&b[i][j]);return 1; }//int inputadjace(int a[][N],int *n) //{ // int i=0,j=0; // printf("點數(shù)：");scanf("%d",n);fflush(stdin); // printf("\n輸入鄰接矩陣的值："); // for(i=0;i<*n;++i) // for(j=0;j<*n;++j) // scanf("%d",&a[i][j]); // fflush(stdin); // return 1; //} // //int inputedge(int b[][N],int n,int *m) //{ // int i=0,j=0; // printf("邊數(shù)：");scanf("%d",m);fflush(stdin); // printf("\n輸入關(guān)聯(lián)矩陣的值："); // for(i=0;i<n;++i) // for(j=0;j<*m;++j) // scanf("%d",&b[i][j]); // return 1; //}int print(int t[][N],int n,int m,char msg[]) {int i=0,j=0;printf("\n%s: \n",msg);for(i=0;i<n;++i){for(j=0;j<m;++j)printf("%4d",t[i][j]);printf("\n");}return 1; }void caldegree(int a[][N],int n,int d[]) {int i=0,j=0;for(i=0;i<n;++i)d[i]=0;for(i=0;i<n;++i)for(j=0;j<n;++j)d[i]+=(a[i][j]==0?0:1); }int iseuler(int d[],int n) {int n0dd=0,i=0;int i0dd1=0,j0dd2=0;printf("\n各點的度數(shù)：");for(i=0;i<n;++i)printf("%4d",d[i]);printf("\n");for(i=0;i<n;++i)if(d[i]%2==1){n0dd++;if(i0dd1==0) i0dd1=i;else j0dd2=i;}if(n0dd==0)printf("\n存在Euler回路，即為Euler圖\n");else if(n0dd==1)printf("\n存在Euler路，起點為%d 終點為%d\n",i0dd1,j0dd2);elseprintf("\n不存在Euler路與回路\n"); }void ishamilton(int d[],int n) {int ishp=1,ishg=1;int i=0,j=0;for(i=0;i<n;++i){for(j=0;j<n;++j)if(d[i]+d[j]<n-1){ishp=0;ishg=0;break;}else if(d[i]+d[j]<n){ishg=0;break;}if(ishp==0) break;}if(ishg==1)printf("為Hamilton圖\n");else if(ishp==1)printf("存在Hamilton路\n");elseprintf("不存在Hamilton路\n"); }void caldegreeb(int b[][N],int n,int m,int d[]) {int i=0,j=0;for(i=0;i<n;++i)d[i]=0;for(i=0;i<n;++i)for(j=0;j<m;++j)if(b[i][j]!=0)d[i]++; }int warshall(int a[][N],int n) {int p[N][N],i=0,j=0,k=0,n1count=0;for(i=0;i<n;++i)for(j=0;j<n;++j)p[i][j]=(a[i][j]>0?1:0);for(j=0;j<n;++j)for(i=0;i<n;++i)if(p[i][j]==1)for(k=0;k<n;++k){p[i][k]=p[i][k]+p[j][k];if(p[i][k]>=1)p[i][k]=1;}for(i=0;i<n;++i)for(j=0;j<n;++j)if(p[i][j]==1)for(k=0;k<n;++k){p[i][k]=p[i][k]+p[j][k];if(p[i][k]>=1)p[i][k]=1;}for(i=0;i<n;++i){for(j=0;j<n;++j){printf("%3d",p[i][j]);n1count+=p[i][j];}printf("\n");}if(n1count==n*n)printf("圖連通\n");elseprintf("圖不連通\n");return n1count; }int poweradjace(int a[][N],int n) {int t[N][N],t2[N][N],aa[N][N],a0[N][N];char msg[100]="長度為",msgi[10];int i=0,j=0,k=0,m=0;for(i=0;i<n;++i)for(j=0;j<n;++j){a0[i][j]=(a[i][j]>0?1:0);t[i][j]=a0[i][j];aa[i][j]=a0[i][j];}print(a0,n,n,strcat(msg,"長度為1的路情況即直接相連的情況"));for(i=2;i<=n;++i){for(j=0;j<n;++j)for(k=0;k<n;++k){t2[j][k]=0;for(m=0;m<n;++m)t2[j][k]=t2[j][k]+t[j][m]*a0[m][k];aa[j][k]==aa[j][k]+t2[j][k];}for(j=0;j<n;++j)for(k=0;k<n;++k)t[j][k]=t2[j][k];itoa(i,msgi,10);strcpy(msg,"長度為");strcat(msg,msgi);strcat(msg,"的路情況");print(t2,n,n,msg);}print(aa,n,n,"匯總表");return 1; }int powellcolor(int a[][N],int n,int color[]) {int d[N],subindex[N],i=0,j=0,k=0,k0=0,itmp=0,thiscolor[N],m=0,nthiscolor=0;for(i=0;i<n;++i){subindex[i]=i;color[i]=0;}caldegree(a,n,d);/* 修改點2：sort排序 */sort(d,d+n);sort(subindex,subindex+n); // for(i=0;i<n-1;++i) // for(j=0;j<n-1-i;++j) // { // if(d[j]<d[j+1]) // { // swap(d[j],d[j+1]); // swap(subindex[j],subindex[j+1]); // } // }printf("排序后的結(jié)點度數(shù):\n");for(i=0;i<n;++i)printf("%4d[%1d]",d[i],subindex[i]);itmp=0;for(i=0;i<n;++i){for(j=0;j<n;++j)thiscolor[j]=0;nthiscolor=0;j=0;while(d[j]==-1&&j<n)j++;if(j>=n)break;k0=subindex[j];itmp++;color[k0]=itmp;d[j]=-1;thiscolor[nthiscolor]=k0;printf("\nk0=%d j=%d nthiscolor=%d itmp=%d\n",k0,j,nthiscolor,itmp);nthiscolor++;j++;while(1){while(d[j]==-1&&j<n)j++;if(j==n)break;k=subindex[j];for(m=0;m<nthiscolor;++m)if(a[k][thiscolor[m]]>0)break;printf("j=%d n=%d k=%d m=%d nthiscolor=%d\n",j,n,k,m,nthiscolor);if(m>=nthiscolor){color[k]=itmp;thiscolor[m]=k;nthiscolor++;d[j]=-1;}j++;}}return 0; }int getedge(int a[][N],int n,stree t[]) {int i=0,j=0,nstree=0;for(i=0;i<n;++i)for(j=i;j<n;++j)if(a[i][j]>0){t[nstree].pointa=i;t[nstree].pointb=j;t[nstree].weight=a[i][j];nstree++;}return nstree; }void printtree(stree t[],int nt) {int i=0;for(i=0;i<nt;++i)printf("%d--%d(%d)",t[i].pointa,t[i].pointb,t[i].weight); }void sortedge(stree t[],int nt) {stree t0;int i=0,j=0;for(i=0;i<nt-1;++i)for(j=0;j<nt-1-i;++j)if(t[j].weight>t[j+1].weight)/* 修改點3：中間元交換值改為swap */swap(t[j],t[j+1]); }int kruskal(stree t0[],int nt0,int n,stree t[]) {int i=0,j=0,k=0,nt=0,b[N][N],m=0,mtmp=0,ncount=0;t[nt]=t0[0];nt++;for(i=1;i<nt0;++i){for(j=0;j<n;++j)for(k=0;k<nt0;++k)b[j][k]=0;for(j=0;j<nt;++j){b[t[j].pointa][j]=1;b[t[j].pointb][j]=1;}b[t[i].pointa][j]=1;b[t[i].pointb][j]=1;printf("加入新邊后的關(guān)聯(lián)矩陣\n");for(j=0;j<n;++j){for(k=0;k<=nt;++k)printf("%4d",b[j][k]);printf("\n");}for(k=0;k<=nt;++k){for(j=k;j<n;++j)if(b[j][k]!=0){if(j>k)for(m=0;m<=nt;++m)swap(b[j][m],b[k][m]);break;}for(j=k+1;j<n;++j)if(b[j][k]==1)for(m=k;m<=nt;++m)b[j][m]=b[j][m]-b[k][m]*b[j][k]/b[k][k];}printf("關(guān)聯(lián)矩陣處理以后：\n");for(j=0;j<n;++j){for(k=0;k<=nt;++k)printf("%4d",b[j][k]);printf("\n");}ncount=0;for(k=0;k<=nt;++k)if(b[k][k]!=0)ncount++;if(ncount==nt+1){t[nt]=t0[i];nt++;if(nt==n-1)break;}}return nt; }int prim(stree t0[],int nt0,int n,stree t[]) {int i=0,j=0,k=0,nt=0,b[N][N],m=0,mtmp=0,ncount=0;int u[N],mdis=0,idis=0,jtree=0,ntree=0;u[nt]=0;printf("u[%d]=%d\n",nt,u[nt]);nt++;while(nt<n){mdis=999999;idis=-1;for(i=0;i<nt;++i){k=u[i];for(j=0;j<nt0;++j){if(t0[j].pointa==k)if(mdis>t0[j].weight){mdis=t0[j].weight;idis=t0[j].pointb;jtree=j;}else if(t0[j].pointb==k)if(mdis>t0[j].weight){mdis=t0[j].weight;idis=t0[j].pointa;jtree=j;}}}u[nt]=idis;t[ntree]=t0[jtree];t0[jtree].weight=999999;printf("nt=%d u[%d]=%d ntree=%d jtree=%d\n",nt,nt,u[nt],ntree,jtree);ntree++;nt++;}return ntree; }int main() {stree st0[N],st1[N];int A[N][N],B[N][N],D[N],color[N];int i=0,j=0,n=0,m=0;int nchoice,nt0=0,nt=0;/* 修改點4：優(yōu)化交互 */bool flag1=false;while(1){printf("\n===========================\n");if(!flag1)printf("1.輸入鄰接矩陣及關(guān)聯(lián)矩陣\n");if(flag1){printf("2.鄰接矩陣求每點度數(shù)及短短圖形性質(zhì)\n3.關(guān)聯(lián)矩陣求每點度數(shù)及判斷圖形性質(zhì)\n4.用warshall算法判斷是否連通\n");printf("5.計算A,A^2,A^3,...\n6.用Powell染色算法對結(jié)點染色\n");printf("7.用kruskal求最小生成樹\n8.用prim求最小生成樹\n");}printf("\n===========================\n");printf("您的選擇(輸入0結(jié)束)：\n");/* 修改點5：判斷輸入是否合法 */while(scanf("%d",&nchoice))if((nchoice>=0&&nchoice<=8)) break;else printf("請重新輸入\n");//scanf("%d",&nchoice);fflush(stdin);if(nchoice==0) break;switch(nchoice){case 1:{inputadjaceedge(A,B,&n,&m);print(A,n,n,"鄰接矩陣");flag1=true;break;} // case 11: // { // inputedge(B,n,&m); // print(B,n,m,"關(guān)聯(lián)矩陣"); // flag11=true; // break; // }case 2:{caldegree(A,n,D);iseuler(D,n);ishamilton(D,n);break;}case 3:{caldegreeb(B,n,m,D);iseuler(D,n);ishamilton(D,n);break;}case 4:{warshall(A,n);break;}case 5:{poweradjace(A,n);break;}case 6:{powellcolor(A,n,color);printf("各點的顏色:\n");for(i=0;i<n;++i)printf("%2d",color[i]);break;}case 7:{nt0=getedge(A,n,st0);printtree(st0,nt0);sortedge(st0,nt0);printtree(st0,nt0);nt=kruskal(st0,nt0,n,st1);printtree(st1,nt);break;}case 8:{nt0=getedge(A,n,st0);printtree(st0,nt0);sortedge(st0,nt0);printtree(st0,nt0);nt=prim(st0,nt0,n,st1);printtree(st1,nt);break;}}}return 0; }

實驗七

/** 實驗七* by ljkjk*//** 共 6 個修改點* 修改點1：萬能頭文件* 修改點2：合并if和for語句，使代碼簡潔易懂* 修改點3：用strchr函數(shù)來代替搜索過程* 修改點4：使用memset函數(shù)替代雙重循環(huán)初始化數(shù)組* 修改點5：利用布爾函數(shù)優(yōu)化交互* 修改點6：利用while判斷輸入是否合法*//* 修改點1：萬能頭文件 */ #include <bits/stdc++.h> #define N 100//集合的基本運算交、并、差、對稱差、直積 int printYsh(char a[]) {//由于集合保存在一個字符串，顯示時在各元素之間加入逗號，及返回長度int nLen = 0;int i = 0;nLen = strlen(a);if (nLen > 0){printf("%c", a[0]);}for (i = 1; i < nLen; i++){printf(",%c", a[i]);}printf("\n");return nLen; } int printRelaYsh(char a[][3], int n) {//由于集合保存在一個字符串，顯示時在各元素之間加入逗號，及返回長度int i = 0; // if (n > 0) // { // printf("<%c,%c>", a[0][0], a[0][1]); // } // for (i = 1; i < n; i++) // { // printf(",<%c,%c>", a[i][0], a[i][1]); // }for (i = 1; i <= n; i++){printf(",<%c,%c>", a[i-1][0], a[i-1][1]);}/* 修改點2：合并if和for語句，使代碼簡潔易懂 */printf("\n");return n; } int printRelaMatrix(int M[][N], int n) {//打印關(guān)系矩陣的值int i = 0, j = 0;for (i = 0; i < n; i++){for (j = 0; j < n; j++)printf("%4d", M[i][j]);printf("\n");} } int inputYsh(char a[]) {//由于集合保存在一個字符串，但是輸入時各字符是用逗號分隔的//形參a是一個數(shù)組，送入是一個地址值，在函數(shù)對該數(shù)組的修改會反映在主函數(shù)char nLen = 0, i = 0, j = 0, k = 0;char stmp[1024];//最多1024個字符printf("集合元素只能是A-Za-z0-9,其他字符被當作分隔符而去掉：\n");gets(stmp);fflush(stdin);nLen = strlen(stmp);for (i = 0; i < nLen; i++){if (((stmp[i] >= 'A') && (stmp[i] <= 'Z')) ||((stmp[i] >= 'a') && (stmp[i] <= 'z')) || ((stmp[i] >= '0') && (stmp[i] <= '9')))//還要判斷該字符 是否已經(jīng)在a中出現(xiàn)過/* 修改點3：用strchr函數(shù)來代替搜索過程 */if(strchr(a,stmp[i])==NULL){a[j]=stmp[i];j++;}}a[j] = '\0';return j;//字符串的長度 } int inputRelaYsh(char a[][3]) {//由于序偶是保存在一個字符串，但是輸入時各字符是用逗號分隔的char nLen = 0, i = 0, j = 0;int nbit = 0;char stmp[1024];//最多1024個字符printf("序偶只能是A-Za-z0-9。其他字符被當作分隔符而去掉，形如<a,b>:\n");gets(stmp);fflush(stdin);nLen = strlen(stmp);for (i = 0; i < nLen; i++){if (((stmp[i] >= 'A') && (stmp[i] <= 'Z')) ||((stmp[i] >= 'a') && (stmp[i] <= 'z')) ||((stmp[i] >= '0') && (stmp[i] <= '9'))){a[j][nbit] = stmp[i];nbit++;if (nbit == 2){a[j][2] = '\0';j++;nbit = 0;}}}return j;//字符串的長度 }int relaCompose(char R[][3], char S[][3], char T[][3], int nR, int nS) {//關(guān)系的復(fù)合，讓RS中每個序偶進行比較，組合成功的寫入到T中int i = 0, j = 0, k = 0, m = 0;char a1 = ' ', a2 = ' ';for (i = 0; i < nR; i++){for (j = 0; j < nS; j++){if (R[i][1] == S[j][0]) //<a,b> of R,<b,c> of S then <a,c> in T{a1 = R[i][0];//還要在T中尋找新序偶是否存在a2 = S[j][1];m = 0;for (m = 0; m < k; m++){//如果已經(jīng)存在則中止if ((T[m][0] == a1) && (T[m][1] == a2)){break;}}if (m >= k) //如果原來不存在則加入其中{T[k][0] = a1;T[k][1] = a2;T[k][2] = '\0';k++;}}}}return k; } int relaSelf(char a[], char R[][3], int nR, char T[][3]) {//自反閉包：判斷集合a中各序偶是否在關(guān)系R中出現(xiàn)，若沒出現(xiàn)則加入其中int i = 0, j = 0, k = 0;int nLen = 0;nLen = strlen(a);//先將關(guān)系R全部轉(zhuǎn)抄到T中for (i = 0; i < nR; i++){T[i][0] = R[i][0];T[i][1] = R[i][1];}for (i = 0; i < nLen; i++){for (j = 0; j < nR; j++){if ((R[j][0] == a[i]) && (R[j][1] == a[i])) //<x,x> in R{break;}}if (j >= nR) //不在關(guān)系R中{T[j][0] = a[i];T[j][1] = a[i];T[j][2] = '\0';nR++;}}return nR; }int relaSym(char R[][3], char T[][3], int nR) {//自反閉包:判斷集合a中各序偶是否在關(guān)系R中出現(xiàn)，若沒出現(xiàn)加入其中int i = 0, j = 0, k = 0;int nLen = nR;//先將關(guān)系R全部轉(zhuǎn)抄到T中for (i = 0; i < nR; i++){T[i][0] = R[i][0];T[i][0] = R[i][0];}for (i = 0; i < nR; i++){for (j = 0; j < nR; j++){//每個序偶到本關(guān)系中尋找對偶關(guān)系，若找到了則是對稱，否則加上if ((R[j][0] == R[i][1]) && (R[j][1] == R[i][0])) //<x,y>,<y,a> inR{break;}}if (j >= nR) //不是成對出現(xiàn){T[nLen][0] = R[i][1];T[nLen][1] = R[i][0];T[nLen][2] = '\0';nLen++;}}return nLen; }void rela2matrix(char a[], char R[][3], int nR, int M[][N]) {//關(guān)系矩陣的行數(shù)與列數(shù)，等于基本元素即數(shù)組a中當前元素的個數(shù)int i = 0, j = 0, na = 0, k = 0, m = 0;na = strlen(a); //基本元素的個數(shù)//關(guān)系矩陣的每個元素清0 // for (i = 0; i < na; i++) // { // for (j = 0; j < na; j++) // { // M[i][j] = 0; // } // }memset(M,0,sizeof(M));/* 修改點4：使用memset函數(shù)替代雙重循環(huán)初始化數(shù)組 *///尋找關(guān)系R中每個序偶之元素對應(yīng)的序號，//從而在數(shù)組M的指定位置置成1for (i = 0; i < nR; i++){//確定第i個序偶的首個元素在基本元素集中的序號for (j = 0; j < na; j++){if (R[i][0] == a[j]){k = j;break;}}//確定第i個序偶的次個元素在基本元素集中的序號for (j = 0; j < na; j++){if (R[i][1] == a[j]){m = j;break;}}M[k][m] = 1;} }int matrix2rela(char a[], char R[][3], int M[][N]) {int i = 0, j = 0, k = 0, nLen = 0;nLen = strlen(a);for (i = 0; i < nLen; i++){for (j = 0; j < nLen; j++){if (M[i][j] == 1){//有次序偶則生成也R[k][0] = a[i];R[k][1] = a[j];k++;}}}return k; } void matrixmulti(int M[][100], int R[][N], int T[][N], int n) {//兩個方程相乘，超過二則為1int i = 0, j = 0, k = 0, nsum = 0;for (i = 0; i < n; i++){for(j=0; j<n; j++){nsum = 0;for (k = 0; k < n; k++){nsum = nsum + M[i][k] * R[k][j];if (nsum >= 1){nsum = 1;break;}}T[i][j] = nsum;}} }int trorder(char R[][3], char RM[][3], int nR, int n) {char T[1024][3], T2[1024][3];//關(guān)系R的i次方int i = 0, nT = 0, j = 0, k = 0, nRM = 0, nT2 = 0;//先將R保存到RM中for (i = 0; i < nR; i++){RM[i][0] = R[i][0];RM[i][1] = R[i][1];T[i][0] = R[i][0];T[i][1] = R[i][1];}nRM = nR;nT = nR;for (i = 2; i <= n; i++)//n-1為止{//將R@R的復(fù)合函數(shù)暫存到TprintRelaYsh(T, nT);printRelaYsh(R, nR);nT2 = relaCompose(T, R, T2, nT, nR);printRelaYsh(T2, nT2);printf("\n");//將復(fù)合以后的添加到R中，得到R+R@R，剔除重復(fù)的元素for (j = 0; j < nT2; j++){for (k = 0; k < nRM; k++){if ((T2[j][0] == RM[k][0]) && (T2[j][1] == RM[k][1])){break;}}if (k >= nRM)//如果沒有找到，則加入其中{RM[nRM][0] = T2[j][0];RM[nRM][1] = T2[j][1];nRM++;}}//將T2復(fù)制給Tfor (j = 0; j < nT2; j++){T[j][0] = T2[j][0];T[j][1] = T2[j][1];}nT = nT2;}return nRM; }int rPower(char R[][3], char T[][3], int nR, int n, int np) {//R的np次方char T2[1024][3];//關(guān)系R的i次方int i = 0, nT = 0, j = 0, k = 0, nRM = 0, nT2 = 0;//先將R保存到T中for (i = 0; i < nR; i++){T[i][0] = R[i][0];T[i][1] = R[i][1];}nT = nR;for (i = 2; i <= np; i++) //n-1為止{//將T@R的復(fù)合暫存到T2nT2 = relaCompose(T, R, T2, nT, nR);//將T2復(fù)制給Tfor (j = 0; j < nT2; j++){T[j][0] = T2[j][0];T[j][1] = T2[j][1];}nT = nT2;}return nT; }int trmatrix(int R[][N], int n) {int i = 0, j = 0, k = 0;for (j = 0; j < n; j++){for (i = 0; i < n; i++){if (R[i][j] == 1){for (k = 0; k < n; k++){//將第j行加到第i行R[i][k] = R[i][k] + R[j][k];if (R[i][k] >= 1){R[i][k] = 1;}}}}}return n; }int main() {char a[1024], R1[1024][3], R2[1024][3], T[1024][3];int nLen1 = 0, nLen2 = 0, nT = 0, M[N][N], nChoice = 0;int MR1[N][N], MR2[N][N];int i = 0, n = 0;/* 修改點5：利用布爾函數(shù)優(yōu)化交互 */bool flag=false;while (1){printf("\n========================\n");if(!flag)printf("1...輸入相關(guān)元素值\n");if(flag){printf("2...R1*R2關(guān)系的復(fù)合\n");printf("3...自反閉包\n");printf("4...對稱閉包\n");printf("5...序偶形式的關(guān)系轉(zhuǎn)關(guān)系矩陣\n");printf("6...利用矩陣求關(guān)系的復(fù)合\n");printf("7...利用序偶形式的復(fù)合求傳遞閉包\n");printf("8...利用warshall算法的求傳遞閉包\n");printf("9...R2*R1關(guān)系的復(fù)合\n");printf("10..R1的任意次方\n");printf("11..R2的任意次方\n");}printf("0...退出\n");printf("========================\n您的選擇: ");//scanf("%d", &nChoice);//fflush(stdin);/* 修改點6：利用while判斷輸入是否合法 */while(scanf("%d",&nChoice))if((nChoice>=0&&nChoice<=8)) break;else printf("請重新輸入\n");if (nChoice == 0){break;}switch (nChoice){case 1:{flag=true;printf("輸入集合a:");inputYsh(a);printf("輸入關(guān)系R1:");nLen1 = inputRelaYsh(R1);printf("輸入關(guān)系R2:");nLen2 = inputRelaYsh(R2);printf("A:");printYsh(a);printf("關(guān)系R1:");printRelaYsh(R1, nLen1);printf("關(guān)系R2:");printRelaYsh(R2, nLen2);break;}case 2:{nT = relaCompose(R1, R2, T, nLen1, nLen2);printf("R1@R2:");printRelaYsh(T, nT);break;}case 9:{nT = relaCompose(R2, R1, T, nLen2, nLen1);printf("R2@R1:");printRelaYsh(T, nT);break;}case 3:{nT = relaSelf(a, R1, nLen1, T);printf("R1的自反閉包：");printRelaYsh(T, nT);break;}case 4:{nT = relaSym(R1, T, nLen1);printf("R1的對稱閉包：");printRelaYsh(T, nT);break;}case 5:{//序偶形式的關(guān)系轉(zhuǎn)換為矩陣形式rela2matrix(a, R1, nLen1, M);printf("R1關(guān)系的序偶");printRelaYsh(R1, nLen1);printf("R1關(guān)系的矩陣\n");printRelaMatrix(M, strlen(a));break;}case 6:{//利用矩陣求關(guān)系的復(fù)合//先求出二個關(guān)系的關(guān)系矩陣rela2matrix(a, R1, nLen1, MR1);printf("R1關(guān)系的矩陣\n");printRelaMatrix(MR1, strlen(a));rela2matrix(a, R2, nLen2, MR2);printf("R2關(guān)系的矩陣\n");printRelaMatrix(MR2, strlen(a));matrixmulti(MR1, MR2, M, strlen(a));printf("復(fù)合后的矩陣\n");printRelaMatrix(M, strlen(a));printf("\n復(fù)合后的序偶");nT = matrix2rela(a, T, M);printRelaYsh(T, nT);break;}case 7://trorder(char R[][3],char RM[][3],int nR,int n){nT = trorder(R1, T, nLen1, strlen(a));printf("R1的傳遞閉包:");printRelaYsh(T, nT);printf("\n序偶轉(zhuǎn)化為數(shù)組\n");rela2matrix(a, T, nT, M);printRelaMatrix(M, strlen(a));break;}case 8:{rela2matrix(a, R1, nLen1, M);printf("R1關(guān)系的序偶");printRelaYsh(R1, nLen1);printf("R1關(guān)系中的矩陣\n");printRelaMatrix(M, strlen(a));trmatrix(M, strlen(a));printf("warshall后的矩陣\n");printRelaMatrix(M, strlen(a));printf("\n序偶");nT = matrix2rela(a, T, M);printRelaYsh(T, nT);break;}case 10:{printf("n=?");scanf("%d", &n);fflush(stdin);nT = rPower(R1, T, nLen1, strlen(a), n);printf("R1的%d次方：", n);printRelaYsh(T, nT);printf("\n序偶轉(zhuǎn)換為數(shù)組\n");rela2matrix(a, T, nT, M);printRelaMatrix(M, strlen(a));break;}case 11:{printf("n=?");scanf("%d", &n);fflush(stdin);nT = rPower(R2, T, nLen2, strlen(a), n);printf("R2的%d次方：", n);printRelaYsh(T, nT);printf("\n序偶轉(zhuǎn)換為數(shù)組\n");rela2matrix(a, T, nT, M);printRelaMatrix(M, strlen(a));break;}//序偶形式的關(guān)系到關(guān)系矩陣/*a,b,c,d,e<a,b>;<b,c>;<a,d>;<c,e>;<e,a>;<d,b><b,a>;<c,b>;<d,a>;<e,c>;<a,e>;<b,d>*/}} }

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