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[導(dǎo)讀]題目鏈接：http://acm.hdu.edu.cn/showproblem.php?pid=5071 題面： Chat Time Limit: 2000/1000 MS (Java/Others

題目鏈接：http://acm.hdu.edu.cn/showproblem.php?pid=5071

題面：

Chat Time Limit: 2000/1000 MS (Java/Others)????Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 2195????Accepted Submission(s): 465

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

解題：

??? 之前一直覺得很煩，沒敢寫，沒想到居然能1A，還是比較驚喜的！

??? 對應(yīng)8種操作，我采用的數(shù)據(jù)結(jié)構(gòu)是兩個map。第一個map為<(int)pos，(node)n>，其中pos為位置，n為節(jié)點，節(jié)點中包含兩個值u（優(yōu)先級權(quán)值），w（單詞數(shù)量）。第二個map為<(int)u,(int)pos>其中u為優(yōu)先級權(quán)值，pos為位置，topp存儲的是被top的那個對象的優(yōu)先級，如果為-1，則說明，沒有對象處在top狀態(tài)，sz為當前節(jié)點數(shù)量。topp和sz都需要時時更新。

??? 1.添加操作，先看是否已經(jīng)存在（map.count判斷），若沒有，則添加，否則顯示錯誤信息。

??? 2.關(guān)閉操作，看是否存在，若不存在，顯示錯誤信息，否則刪除當前節(jié)點，并將后面的往上順移一位。

??? 3.聊天操作，看sz是否為0，若為0，則顯示錯誤信息。若不為0，則看topp是否為-1，若不為-1，則通過map2找到對應(yīng)位置，更新map1中的值，若為-1，則更新位置為1的map1中的值。

??? 4.置頂操作，首先判斷是否越界，不越界則將該位置提到首位即可，其實應(yīng)該寫成函數(shù)供下面功能使用，不過好在代碼都是差不多的，拷貝下就好了。

??? 5.優(yōu)先級置頂，循環(huán)遍歷，然后將該位置置頂，與上同。

??? 6.通過map2找到對應(yīng)位置，并置頂。

??? 7.將topp值改成此時的u值。

??? 8.將topp還原為-1。

??? 最后要注意還要輸出和剩下的沒關(guān)閉的窗口聊天的數(shù)量，若為0，則不需輸出，同時要將優(yōu)先級最高的先輸出。

坑點：

??? 1.topp值和sz值需時時維護。

??? 2.數(shù)量應(yīng)用long long保存。

??? 3.輸出答案時，最后是有'.'的。

??? 4.最后輸出窗口要先輸出處于top狀態(tài)的，沒有就按順序即可。

代碼：

#include 
#include 
#include 
#include 
#define LL long long 
using namespace std;
struct node
{
   node (int x,LL y)
   {
	   u=x;
	   w=y;
   }
   node ()
   {
	   u=0;
	   w=0;
   }
   node (const node &xx)
   {
	   u=xx.u;
	   w=xx.w;
   }
   //u優(yōu)先級權(quán)值，w數(shù)量
   int u;
   LL w;
};
map  store;
map  refl;
int main()
{
    int t,n,val,sz,pos,topp,uu,maxn;
	LL amt,ww;
	char oper[10];
	scanf("%d",&t);
	node tmp;
	for(int i=1;i<=t;i++)
	{
		store.clear();
		refl.clear();
		sz=0;
		topp=-1;
		scanf("%d",&n);
		for(int j=1;j<=n;j++)
		{
			printf("Operation #%d: ",j);
			getchar();
			scanf("%s",oper);
			//加操作
			if(oper[0]=='A')
			{
				scanf("%d",&val);
                if(!refl.count(val))
				{
					printf("success.n");
                    tmp.w=0;
					tmp.u=val;
                    pos=++sz;
                    refl[val]=pos;
					store[pos]=tmp;
				}
				else
				printf("same priority.n");
			}
			//關(guān)閉操作
			else if(oper[0]=='C'&&oper[1]=='l')
			{
				scanf("%d",&val);
				if(!refl.count(val))
				  printf("invalid priority.n");
				else
				{
                  if(val==topp)
					  topp=-1;
                  pos=refl[val];
				  amt=store[pos].w;
                  refl.erase(val);
				  for(int k=pos+1;k<=sz;k++)
				  {
					  uu=store[k].u;
					  store[k-1].u=store[k].u;
					  store[k-1].w=store[k].w;
					  refl[uu]=k-1;
				  }
				  store.erase(sz);
				  sz--;
                  printf("close %d with %d.n",val,amt);
				}
			}
			//聊天功能
			else if(oper[0]=='C'&&oper[2]=='a')
			{
				scanf("%d",&val);
				if(sz==0)
					printf("empty.n");
				else
				{
					printf("success.n");
					if(topp==-1)
						store[1].w+=val;
					else
					{
						pos=refl[topp];
						store[pos].w+=val;
					}
				}
			}
			//置頂操作
			else if(oper[0]=='R')
			{
				scanf("%d",&val);
				if(val<1||val>sz)
					printf("out of range.n");
				else
				{
				   printf("success.n");
                   uu=store[val].u;
				   ww=store[val].w;
				   for(int k=val-1;k>=1;k--)
				   {
                     store[k+1].u=store[k].u;
					 store[k+1].w=store[k].w;
					 refl[store[k+1].u]=k+1;
				   }
				   store[1].u=uu;
				   store[1].w=ww;
				   refl[uu]=1;
				}
			}
			//優(yōu)先級置頂
			else if(oper[0]=='P')
			{
               if(sz==0)
				   printf("empty.n");
			   else
			   {
				   printf("success.n");
				   maxn=0;
				   for(int k=1;k<=sz;k++)
				   {
					   if(store[k].u>maxn)
						   maxn=store[k].u;
				   }
				   val=refl[maxn];
				   uu=store[val].u;
				   ww=store[val].w;
				   for(int k=val-1;k>=1;k--)
				   {
                     store[k+1].u=store[k].u;
					 store[k+1].w=store[k].w;
					 refl[store[k+1].u]=k+1;
				   }
				   store[1].u=uu;
				   store[1].w=ww;
				   refl[uu]=1;
			   }
			}
			//選擇置頂
			else if(oper[0]=='C')
			{
				scanf("%d",&val);
				if(!refl.count(val))
			      printf("invalid priority.n");
				else
				{
					printf("success.n");
					val=refl[val];
 				    uu=store[val].u;
				    ww=store[val].w;
				    for(int k=val-1;k>=1;k--)
				    {
                      store[k+1].u=store[k].u;
					  store[k+1].w=store[k].w;
					  refl[store[k+1].u]=k+1;
				    }
				    store[1].u=uu;
				    store[1].w=ww;
				    refl[uu]=1;
				}
			}
			//top操作
			else if(oper[0]=='T')
			{
				scanf("%d",&val);
				if(!refl.count(val))
                  printf("invalid priority.n");
				else
				{
					printf("success.n");
					topp=val;
				}
			}
			//untop操作
			else if(oper[0]=='U')
			{
                if(topp==-1)
					printf("no such person.n");
				else
				{
					printf("success.n");
					topp=-1;
				}
			}
		}
		//特殊優(yōu)先級優(yōu)先
		if(topp!=-1&&store[refl[topp]].w)
			printf("Bye %d: %I64dn",topp,store[refl[topp]].w);
		for(int i=1;i<=sz;i++)
		{
			//優(yōu)先級只需輸出一次
			if(store[i].u!=topp&&store[i].w>0)
				printf("Bye %d: %I64dn",store[i].u,store[i].w);
		}
	}
	return 0;
}

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