Chat

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)

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

Hint

This 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

题意：

就是要你写一个聊天窗口管理系统。支持增加窗口。关闭窗口。管理优先级。。。。。详细的还是自己读吧。

思路：

觉得splay可做就用splay做的。功能也是splay最基本的功能。这题需要注意的是。最后Bye的时候是先给置顶的人然后按队列顺序Bye。

详细见代码：

#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];//子结点的和
map<int,int> mp;
int sz[maxn<<1],pre[maxn<<1],mv[maxn<<1],prr[maxn<<1],ch[maxn<<1][2];//sz记录子树规模,pre记录父结点,val存结点值。ch[k][0],ch[k][1]k的左右儿子。
int val[maxn<<1];
int mi,mo,root,s[maxn<<1];//mi回收内存。mo分配内存。root为根。s为内存池
void Treaval(int x)//Debug部分打出来非常清楚
{
    if(x)
    {
        Treaval(ch[x][0]);
        printf("结点%2d:左儿子 %2d 右儿子 %2d 父结点 %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记录可用空间大小但里面没有空间所以要先减减。不然会出错
        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()//初始化很重要！
{
    mi=root=0;
    mo=1;
    ch[0][0]=ch[0][1]=pre[0]=sz[0]=mv[0]=prr[0]=0;
    val[0]=0;//建一个虚拟根节点。做标记用。pre==0说明就到根了
    Newnode(root,0,0,0);//建真正的根和右儿子。必须加这两个虚拟结点。这样区间操作起来才方便
    Newnode(ch[root][1],0,0,root);
    PushUp(ch[root][1]);
    PushUp(root);
}
void Rotate(int x,int w)//旋转。w为旋转方式。0为ZAG(x在右边)。1为ZIG(x在左边)。结点为左子树就右旋。右子树就左旋
{
    int y=pre[x];
    ch[y][!w]=ch[x][w];//把x往上转
    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为目标父结点
{
    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)//旋转到根时更换根结点
        root=rt;
}
int Get_kth(int rt,int k)//取第k个数的标号
{
    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位置插入一个数
{
    Splay(Get_kth(root,pos),0);//因为前面有一个虚拟结点所以实际插的位置为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)//删除结点
{
    Splay(rt,0);//先把该点旋转到根
    int pos=sz[ch[rt][0]];//获取它前面有多少个数
    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;
}