Chat

Time Limit: 2000/1000 MS (Java/Others)????Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 571????Accepted Submission(s): 136


Problem Description As everyone knows, DRD has no girlfriends. But as everyone also knows, DRD’s friend ATM’s friend CLJ has many potential girlfriends. One evidence is CLJ’s chatting record.


CLJ chats with many girls all the time. Sometimes he begins a new conversation and sometimes he ends a conversation. Sometimes he chats with the girl whose window is on the top.

You can imagine CLJ’s windows as a queue. The first girl in the queue is the top girl if no one is “always on top ”.

Since CLJ is so popular, he begins to assign a unique positive integer as priority for every girl. The higher priority a girl has, the more CLJ likes her. For example, GYZ has priority 10⁹, and JZP has priority 10⁸ while Sister Soup has priority 1, and Face Face has priority 2.

As a famous programmer, CLJ leads a group to implement his own WM(window manager). The WM will log CLJ’s operations. Now you are supposed to implement the log system. The general logging format is “Operation #X: LOGMSG.”, where X is the number of the operation and LOGMSG is the logging message.

There are several kinds of operations CLJ may use:

1.Add u: CLJ opens a new window whose priority is u, and the new window will be the last window in the window queue. This operation will always be successful except the only case in which there is already a window with priority u. If it is successful, LOGMSG will be “success”. Otherwise LOGMSG will be “same priority”.

2.Close u: CLJ closes a window whose priority is u. If there exists such a window, the operation will be successful and LOGMSG will be “close u with c”, where u is the priority and c is the number of words CLJ has spoken to this window. Otherwise, LOGMSG will be “invalid priority”. Note that ANY window can be closed.

3.Chat w: CLJ chats with the top window, and he speaks w words. The top window is the first window in the queue, or the “always on top” window (as described below) instead if there exists. If no window is in the queue, LOGMSG will be “empty”, otherwise the operation can be successful and LOGMSG will be “success”.

4.Rotate x: CLJ performs one or more Alt-Tabs to move the x-th window to the first one in the queue. For example, if there are 4 windows in the queue, whose priorities are 1, 3, 5, 7 respectively and CLJ performs “Rotate 3”, then the window’s priorities in the queue will become 5, 1, 3, 7. Note that if CLJ wants to move the first window to the head, this operation is still considered “successful”. If x is out of range (smaller than 1 or larger than the size of the queue), LOGMSG will be “out of range”. Otherwise LOGMSG should be “success”.

5.Prior: CLJ finds out the girl with the maximum priority and then moves the window to the head of the queue. Note that if the girl with the maximum priority is already the first window, this operation is considered successful as well. If the window queue is empty, this operation will fail and LOGMSG must be “empty”. If it is successful, LOGMSG must be “success”.

6.Choose u: CLJ chooses the girl with priority u and moves the window to the head of the queue.This operation is considered successful if and only if the window with priority u exists. LOGMSG for the successful cases should be “success” and for the other cases should be “invalid priority”.

7.Top u: CLJ makes the window of the girl with priority u always on top. Always on top is a special state, which means whoever the first girl in the queue is, the top one must be u if u is always on top. As you can see, two girls cannot be always on top at the same time, so if one girl is always on top while CLJ wants another always on top, the first will be not always on top any more, except the two girls are the same one. Anyone can be always on top. LOGMSG is the same as that of the Choose operation.

8.Untop: CLJ cancels the “always on top” state of the girl who is always on top. That is, the girl who is always on top now is not in this special state any more. This operation will fail unless there is one girl always on top. If it fails, LOGMSG should be “no such person”, otherwise should be “success”.

As a gentleman, CLJ will say goodbye to every active window he has ever spoken to at last, “active” here means the window has not been closed so far. The logging format is “Bye u: c” where u is the priority and c is the number of words he has ever spoken to this window. He will always say good bye to the current top girl if he has spoken to her before he closes it.
Input The first line contains an integer T (T ≤ 10), denoting the number of the test cases.

For each test case, the first line contains an integer n(0 < n ≤ 5000), representing the number of operations. Then follow n operations, one in a line. All the parameters are positive integers below 10⁹.
Output Output all the logging contents.
Sample Input 1 18 Prior Add 1 Chat 1 Add 2 Chat 2 Top 2 Chat 3 Untop Chat 4 Choose 2 Chat 5 Rotate 2 Chat 4 Close 2 Add 3 Prior Chat 2 Close 1
Sample Output Operation #1: empty. Operation #2: success. Operation #3: success. Operation #4: success. Operation #5: success. Operation #6: success. Operation #7: success. Operation #8: success. Operation #9: success. Operation #10: success. Operation #11: success. Operation #12: success. Operation #13: success. Operation #14: close 2 with 8. Operation #15: success. Operation #16: success. Operation #17: success. Operation #18: close 1 with 11. Bye 3: 2 HintThis problem description does not relate to any real person in THU.
Source 2014 Asia AnShan Regional Contest
Recommend liuyiding???|???We have carefully selected several similar problems for you:??5081?5080?5079?5078?5077? 題意： 就是要你寫一個聊天窗體管理系統。支持添加窗體。

關閉窗體。管理優先級。。

。。。具體的還是自己讀吧。

思路： 認為splay可做就用splay做的。功能也是splay最主要的功能。這題須要注意的是。最后Bye的時候是先給置頂的人然后按隊列順序Bye。 具體見代碼： #include<iostream> #include<stdio.h> #include<string.h> #include<algorithm> #include<map> using namespace std; typedef long long ll; const int maxn=100100; const int INF=0x3f3f3f3f; //ll sum[maxn<<1];//子結點的和 map<int,int> mp; int sz[maxn<<1],pre[maxn<<1],mv[maxn<<1],prr[maxn<<1],ch[maxn<<1][2];//sz記錄子樹規模,pre記錄父結點,val存結點值。

ch[k][0],ch[k][1]k的左右兒子。

int val[maxn<<1]; int mi,mo,root,s[maxn<<1];//mi回收內存。mo分配內存。

root為根。s為內存池 void Treaval(int x)//Debug部分打出來非常清楚 { if(x) { Treaval(ch[x][0]); printf("結點%2d:左兒子 %2d 右兒子 %2d 父結點 %2d size = %2d ,val = %2d \n",x,ch[x][0],ch[x][1],pre[x],sz[x],val[x]); Treaval(ch[x][1]); } } void Debug() { printf("root:%d\n",root); Treaval(root); } void Newnode(int &rt,int v,int pr,int fa) { if(mi)//注意mi記錄可用空間大小但里面沒有空間所以要先減減。不然會出錯 rt=s[--mi]; else rt=mo++; ch[rt][0]=ch[rt][1]=0; //sum[r]=k; pre[rt]=fa; prr[rt]=pr; sz[rt]=1; mv[rt]=rt; val[rt]=v; } void PushUp(int rt) { int ls,rs; if(rt==0) return; ls=ch[rt][0],rs=ch[rt][1]; sz[rt]=sz[ls]+sz[rs]+1; if(prr[mv[ls]]>prr[mv[rs]]) mv[rt]=mv[ls]; else mv[rt]=mv[rs]; if(prr[rt]>prr[mv[rt]]) mv[rt]=rt; //sum[rt]=sum[ls]+sum[rs]+val[rt]; } void Init()//初始化非常重要！ { mi=root=0; mo=1; ch[0][0]=ch[0][1]=pre[0]=sz[0]=mv[0]=prr[0]=0; val[0]=0;//建一個虛擬根節點。做標記用。

pre==0說明就到根了 Newnode(root,0,0,0);//建真正的根和右兒子。必須加這兩個虛擬結點。這樣區間操作起來才方便 Newnode(ch[root][1],0,0,root); PushUp(ch[root][1]); PushUp(root); } void Rotate(int x,int w)//旋轉。w為旋轉方式。

0為ZAG(x在右邊)。1為ZIG(x在左邊)。

結點為左子樹就右旋。

右子樹就左旋 { int y=pre[x]; ch[y][!w]=ch[x][w];//把x往上轉 pre[ch[x][w]]=y; //PushDown(y) //PushDown(x) if(pre[y]) ch[pre[y]][ch[pre[y]][1]==y]=x; pre[x]=pre[y]; ch[x][w]=y; pre[y]=x; PushUp(y); } void Splay(int rt,int goal)//goal為目標父結點 { int y,w; while(pre[rt]!=goal) { if(pre[pre[rt]]==goal) Rotate(rt,ch[pre[rt]][0]==rt); else { y=pre[rt]; w=(ch[pre[y]][0]==y); if(ch[y][w]==rt) { Rotate(rt,!w); Rotate(rt,w); } else { Rotate(y,w); Rotate(rt,w); } } } PushUp(rt); if(goal==0)//旋轉到根時更換根結點 root=rt; } int Get_kth(int rt,int k)//取第k個數的標號 { int tp=sz[ch[rt][0]]+1; if(tp==k) return rt; if(tp>k) return Get_kth(ch[rt][0],k); else return Get_kth(ch[rt][1],k-tp); } int Insert(int pos,int v,int pr)//在pos位置插入一個數 { Splay(Get_kth(root,pos),0);//由于前面有一個虛擬結點所以實際插的位置為pos+1 Splay(Get_kth(root,pos+1),root); Newnode(ch[ch[root][1]][0],v,pr,ch[root][1]); PushUp(ch[root][1]); PushUp(root); return ch[ch[root][1]][0]; } void Delete(int rt)//刪除結點 { Splay(rt,0);//先把該點旋轉到根 int pos=sz[ch[rt][0]];//獲取它前面有多少個數 Splay(Get_kth(root,pos),0); Splay(Get_kth(root,pos+2),root); s[mi++]=rt; ch[ch[root][1]][0]=0; PushUp(ch[root][1]); PushUp(root); } int main() { int n,op,i,t,pos,rt,top; char cmd[15]; scanf("%d",&t); while(t--) { scanf("%d",&n); Init(); mp.clear(); top=-1; for(i=1;i<=n;i++) { scanf("%s",cmd); if(!strcmp(cmd,"Add")) { scanf("%d",&op); if(mp.count(op)) printf("Operation #%d: same priority.\n",i); else { pos=mp.size(); rt=Insert(pos+1,0,op); mp[op]=rt; printf("Operation #%d: success.\n",i); } } else if(!strcmp(cmd,"Close")) { scanf("%d",&op); if(mp.count(op)) { printf("Operation #%d: close %d with %d.\n",i,op,val[mp[op]]); Delete(mp[op]); if(top==mp[op]) top=-1; mp.erase(op); } else printf("Operation #%d: invalid priority.\n",i); } else if(!strcmp(cmd,"Chat")) { scanf("%d",&op); if(top!=-1) { val[top]+=op; printf("Operation #%d: success.\n",i); } else if(mp.size()) { pos=Get_kth(root,2); val[pos]+=op; printf("Operation #%d: success.\n",i); } else printf("Operation #%d: empty.\n",i); } else if(!strcmp(cmd,"Rotate")) { scanf("%d",&op); if(mp.size()>=op) { pos=Get_kth(root,op+1); Delete(pos); rt=Insert(1,val[pos],prr[pos]); mp[prr[pos]]=rt; if(pos==top) top=rt; printf("Operation #%d: success.\n",i); } else printf("Operation #%d: out of range.\n",i); } else if(!strcmp(cmd,"Prior")) { if(mp.size()) { pos=mv[root]; Delete(pos); rt=Insert(1,val[pos],prr[pos]); mp[prr[pos]]=rt; if(top==pos) top=rt; printf("Operation #%d: success.\n",i); } else printf("Operation #%d: empty.\n",i); } else if(!strcmp(cmd,"Choose")) { scanf("%d",&op); if(mp.count(op)) { pos=mp[op]; Delete(pos); rt=Insert(1,val[pos],prr[pos]); if(top==pos) top=rt; mp[op]=rt; printf("Operation #%d: success.\n",i); } else printf("Operation #%d: invalid priority.\n",i); } else if(!strcmp(cmd,"Top")) { scanf("%d",&op); if(mp.count(op)) { top=mp[op]; printf("Operation #%d: success.\n",i); } else printf("Operation #%d: invalid priority.\n",i); } else if(!strcmp(cmd,"Untop")) { if(top!=-1) { top=-1; printf("Operation #%d: success.\n",i); } else printf("Operation #%d: no such person.\n",i); } } if(top!=-1) { if(val[top]) printf("Bye %d: %d\n",prr[top],val[top]); Delete(top); } while(sz[root]>2) { pos=Get_kth(root,2); if(val[pos]) printf("Bye %d: %d\n",prr[pos],val[pos]); Delete(pos); } } return 0; }

轉載于:https://www.cnblogs.com/blfshiye/p/5174904.html

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