P1541 乌龟棋
題面:https://www.luogu.org/problem/P1541
本題由于每種牌個(gè)數(shù)<=40 所以設(shè)f[a][b][c][d]表示你出了a張爬行牌1,b張爬行牌2,c張爬行牌3,d張爬行牌4時(shí)的得分 然后直接枚舉每種牌的個(gè)數(shù)遞推即可Code: #include<iostream> #include<cstdio> #include<cstring> #include<cstdlib> #include<cmath> #include<algorithm> #include<ctime> #include<queue> using namespace std; const int N=355,M=45; int n,m,a[N],b,f[M][M][M][M],num[N]; int main(){scanf("%d%d",&n,&m);for(int i=1;i<=n;i++){scanf("%d",&a[i]);}for(int i=1;i<=m;i++){scanf("%d",&b);num[b]++;}f[0][0][0][0]=a[1];for(int i=0;i<=num[1];i++){for(int j=0;j<=num[2];j++){for(int k=0;k<=num[3];k++){for(int l=0;l<=num[4];l++){int r=i+j*2+k*3+l*4+1;if(i!=0){f[i][j][k][l]=max(f[i][j][k][l],f[i-1][j][k][l]+a[r]);}if(j!=0){f[i][j][k][l]=max(f[i][j][k][l],f[i][j-1][k][l]+a[r]);}if(k!=0){f[i][j][k][l]=max(f[i][j][k][l],f[i][j][k-1][l]+a[r]);}if(l!=0){f[i][j][k][l]=max(f[i][j][k][l],f[i][j][k][l-1]+a[r]);}}}}}printf("%d\n",f[num[1]][num[2]][num[3]][num[4]]);return 0; }轉(zhuǎn)載于:https://www.cnblogs.com/ukcxrtjr/p/11470703.html
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