大模拟:

1》saygoodbye要先对 always on top 的人说
2》对没有说过话的不要说good bye
3》用long long

Chat

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

Total Submission(s): 337    Accepted Submission(s): 82

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
 

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <set> using namespace std; typedef long long int LL; const int INF=0x3f3f3f3f; const int HEAD=0;
const int TAIL=6400; struct CHART
{
int pro,front,back;
LL w;
}chat[6500]; int n,x,id;
int alwayson;
set<int> pc;
char cmd[50]; void init()
{
id=1;pc.clear();alwayson=-1;
memset(chat,0,sizeof(chat));
chat[HEAD].front=HEAD; chat[HEAD].back=TAIL;
chat[TAIL].front=HEAD; chat[TAIL].back=TAIL;
chat[HEAD].pro=-INF-1; chat[TAIL].pro=-INF-1;
} void Add(int u)
{
if(pc.count(u))
{
puts("same priority.");
return ;
}
pc.insert(u);
chat[id].w=0; chat[id].pro=u;
chat[id].front=chat[TAIL].front;
chat[id].back=TAIL;
chat[chat[TAIL].front].back=id;
chat[TAIL].front=id;
id++;
puts("success.");
} void Close(int u)
{
if(pc.count(u)==0)
{
puts("invalid priority.");
return ;
}
pc.erase(u);
if(alwayson==u) alwayson=-1;
int pos=HEAD;
for(pos=HEAD;pos!=TAIL;pos=chat[pos].back)
{
if(chat[pos].pro==u) break;
}
int bc=chat[pos].back;
int ft=chat[pos].front;
chat[bc].front=ft;
chat[ft].back=bc;
printf("close %d with %I64d.\n",chat[pos].pro,chat[pos].w);
} void Chat(int w)
{
if(chat[HEAD].back==TAIL)
{
puts("empty.");
return ;
}
puts("success.");
int u=-1;
if(alwayson!=-1) u=alwayson;
else u=chat[chat[HEAD].back].pro;
int pos=HEAD;
for(pos=HEAD;pos!=TAIL;pos=chat[pos].back)
{
if(chat[pos].pro==u) break;
}
chat[pos].w+=w;
} void Rotate(int x)
{
if(x<1||x>pc.size())
{
puts("out of range.");
return ;
}
puts("success.");
int pos=HEAD,i=0;
for(pos=HEAD;i<x&&pos!=TAIL;pos=chat[pos].back,i++) ;
///split
int bc=chat[pos].back;
int ft=chat[pos].front;
chat[bc].front=ft;
chat[ft].back=bc;
///merge
chat[pos].front=HEAD;
chat[pos].back=chat[HEAD].back;
chat[chat[HEAD].back].front=pos;
chat[HEAD].back=pos;
} void Prior()
{
if(chat[HEAD].back==TAIL)
{
puts("empty.");
return ;
}
puts("success.");
int pos=HEAD;
int mxp=HEAD;
for(pos=HEAD;pos!=TAIL;pos=chat[pos].back)
{
if(chat[pos].pro>chat[mxp].pro)
{
mxp=pos;
}
}
pos=mxp;
///split
int bc=chat[pos].back;
int ft=chat[pos].front;
chat[bc].front=ft;
chat[ft].back=bc;
///merge
chat[pos].front=HEAD;
chat[pos].back=chat[HEAD].back;
chat[chat[HEAD].back].front=pos;
chat[HEAD].back=pos;
} void Choose(int u)
{
if(pc.count(u)==0)
{
puts("invalid priority.");
return ;
}
puts("success.");
int pos=HEAD;
for(pos=HEAD;pos!=TAIL;pos=chat[pos].back)
{
if(chat[pos].pro==u) break;
}
///split
int bc=chat[pos].back;
int ft=chat[pos].front;
chat[bc].front=ft;
chat[ft].back=bc;
///merge
chat[pos].front=HEAD;
chat[pos].back=chat[HEAD].back;
chat[chat[HEAD].back].front=pos;
chat[HEAD].back=pos;
} void Top(int u)
{
if(pc.count(u)==0)
{
puts("invalid priority.");
return ;
}
puts("success.");
alwayson=u;
} void Untop()
{
if(alwayson==-1)
{
puts("no such person.");
return ;
}
alwayson=-1;
puts("success.");
} void saybyebye()
{
if(alwayson!=-1)
{
int p=HEAD;
for(p=HEAD;p!=TAIL;p=chat[p].back)
{
if(chat[p].pro==alwayson) break;
}
if(chat[p].w) printf("Bye %d: %I64d\n",chat[p].pro,chat[p].w);
}
int pos=HEAD;
for(pos=chat[HEAD].back;pos!=TAIL;pos=chat[pos].back)
{
if(chat[pos].pro!=alwayson&&chat[pos].w)
printf("Bye %d: %I64d\n",chat[pos].pro,chat[pos].w);
}
} int main()
{
int T_T;
scanf("%d",&T_T);
while(T_T--)
{
init();
scanf("%d",&n);
for(int i=0;i<n;i++)
{
scanf("%s",cmd);
printf("Operation #%d: ",i+1);
if(strcmp(cmd,"Add")==0)
{
int u; scanf("%d",&u);
Add(u);
}
else if(strcmp(cmd,"Close")==0)
{
int u; scanf("%d",&u);
Close(u);
}
else if(strcmp(cmd,"Chat")==0)
{
int u; scanf("%d",&u);
Chat(u);
}
else if(strcmp(cmd,"Rotate")==0)
{
int u; scanf("%d",&u);
Rotate(u);
}
else if(strcmp(cmd,"Prior")==0)
{
Prior();
}
else if(strcmp(cmd,"Choose")==0)
{
int u; scanf("%d",&u);
Choose(u);
}
else if(strcmp(cmd,"Top")==0)
{
int u; scanf("%d",&u);
Top(u);
}
else if(strcmp(cmd,"Untop")==0)
{
Untop();
}
}
saybyebye();
}
return 0;
}

HDOJ 5071 Chat 模拟的更多相关文章

  1. hdu 5071 Chat(模拟)

    题目链接:hdu 5071 Chat 题目大意:模拟题. .. 注意最后说bye的时候仅仅要和讲过话的妹子说再见. 解题思路:用一个map记录每一个等级的妹子讲过多少话以及是否有这个等级的妹子.数组A ...

  2. hdu 5071 Chat(模拟|Splay)

    Chat Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others) Total Sub ...

  3. HDU 5071 Chat(2014鞍山B,模拟)

    http://acm.hdu.edu.cn/showproblem.php?pid=5071 Chat Time Limit: 2000/1000 MS (Java/Others)    Memory ...

  4. HDU - 5071 Chat(模拟)

    原题链接 题意:有各种操作,模拟这个程序并输出每次操作的信息 分析:恶心模拟题...用个map记录一下各个等级女孩的谈话数,同时也便于查找权值为u的在不在队列里.因为n很小,其他就暴力模拟了. #in ...

  5. HDU 5071 Chat(2014鞍山赛区现场赛B题)

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5071 解题报告:一个管理聊天窗口的程序,一共有八种操作,然后要注意的就是Top操作只是把编号为u的窗口 ...

  6. HDU 5071 Chat

    题意: CLJ找了很多妹子-  (题目好没节操-)  对于CLJ和妹子的聊天对话框  有一下几种操作: add  加一个妹子在聊天窗队列末尾  假设这个妹子已经在队列中则add失败 close  关掉 ...

  7. hdoj 5319 Painter(模拟题)

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5319 思路分析:假设颜色R表示为1,颜色B表示为2,颜色G表示为3,因为数据量较小,采用暴力解法即可, ...

  8. hdu 5071 Chat-----2014acm亚洲区域赛鞍山 B题

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5071 Chat Time Limit: 2000/1000 MS (Java/Others)    M ...

  9. hdu 5071(2014鞍山现场赛B题,大模拟)

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5071 思路:模拟题,没啥可说的,移动的时候需要注意top的变化. #include <iostr ...

随机推荐

  1. WCF跟踪分析 使用(SvcTraceViewer)

    1.首先在WCF服务端配置文件中配置两处,用于记录WCF调用记录! A:<system.serviceModel>目录下: <diagnostics>      <mes ...

  2. js中访问对象的方法

    如果在js中定义了一个变量obj1,如 var obj1 = 234; 那么访问这个边个两的方式至少有两种, 1 window["obj1"],那么值为234, 2 var tar ...

  3. Active MQ C#实现

    原文链接: Active MQ C#实现 内容概要 主要以源码的形式介绍如何用C#实现同Active MQ 的通讯.本文假设你已经正确安装JDK1.6.x,了解Active MQ并有一定的编程基础. ...

  4. 【转】NP-Hard和NP-Complete的区别

    原文来自:http://hi.baidu.com/nuclearspace/item/e0f8a1b777914974254b09f4 对 NP-Hard问题和NP-Complete问题的一个直观的理 ...

  5. 通过Qt样式表定制程序外观(比较通俗易懂)

    1. 何为Qt样式表[喝小酒的网摘]http://blog.hehehehehe.cn/a/10270.htm2. 样式表语法基础3. 方箱模型4. 前景与背景5. 创建可缩放样式6. 控制大小7. ...

  6. 基于QtQuick2.0应用程序运行于XP系统的诸多问题

    客户端 使用QtQuick技术开发酷炫的XP客户端经常遇到白屏或者无界面 if Qt is built using ANGLE, its shared libraries and the requir ...

  7. js 选择器

    a>b  获取a下面的直接子元素

  8. vs2010 调试中监视变量

    在msdn中写了如何查看调试的数据,网址: http://msdn.microsoft.com/zh-cn/library/vstudio/esta7c62(v=vs.100).aspx Visual ...

  9. 点菜系统 pickview的简单实用

    使用pickview的时候多想想tableview的使用,观察两者的相同之处 pickview的主要用途用于选择地区  生日年月日  和点餐 示例代码 简单的pickview点餐系统//  ViewC ...

  10. VIM: 解决vi/vim中粘贴时行首出现很多缩进和空格的问题

    解决vi/vim中粘贴时行首出现很多缩进和空格的问题 http://www.jbxue.com/LINUXjishu/12232.html 由于在secureCRT中会将原来的文本原封不动的按照字符串 ...