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. MVCC 多版本并发控制

    关于事务的介绍暂且不谈. InnoDB行级锁,虽然在很大程度上提高了事务的并发性,但是终究还是要耗费很大的.为了更进一步的提高并发性同时降低开销,存储引擎会同时实现MVCC. InnoDB实现MVCC ...

  2. jQuery 动态元素添加

    有这么一道题 <!DOCTYPE html> <head> <title>前端工程师面试题</title> <meta http-equiv=&q ...

  3. 利用Mono.Cecil动态修改程序集来破解商业组件(仅用于研究学习)

    原文 利用Mono.Cecil动态修改程序集来破解商业组件(仅用于研究学习) Mono.Cecil是一个强大的MSIL的注入工具,利用它可以实现动态创建程序集,也可以实现拦截器横向切入动态方法,甚至还 ...

  4. Poj 2299 Ultra-QuickSort(归并排序)

    题目链接:http://poj.org/problem?id=2299 思路分析:序列的逆序数即为交换次数,所以求出该序列的逆序数即可.根据分治法思想,序列分为两个大小相等的两部分, 分别求子序列的逆 ...

  5. 探索Android中的Parcel机制(上)

    一.先从Serialize说起 我们都知道JAVA中的Serialize机制,译成串行化.序列化……,其作用是能将数据对象存入字节流其中,在须要时又一次生成对象.主要应用是利用外部存储设备保存对象状态 ...

  6. .Net之路(十三)数据库导出到EXCEL

    .NET中导出到Office文档(word,excel)有我理解的两种方法.一种是将导出的文件存放在server某个目录以下,利用response输出到浏览器地址栏,直接打开:还有直接利用javasc ...

  7. 淘宝PK京东:哥刷的不是广告,刷的是存在

    冯强/文 (昨晚看阿根廷vs瑞士时手机上敲的,看完太激动忘发了,现配了图发上来) 这两天,关于京东.淘宝渠道下沉的新闻中,两家略带喜感的农村墙体广告在互联网上传播,例如以下图: 京东这图片,越看越像P ...

  8. ios中利用NSDateComponents、NSDate、NSCalendar判断当前时间是否在一天的某个时间段内。

    应用中设置一般会存在这样的设置,如夜间勿扰模式,从8:00-23:00,此时如何判断当前时间是否在该时间段内.难点主要在于如何用NSDate生成一个8:00的时间和23:00的时间,然后用当前的时间跟 ...

  9. ThreadLocal 和 InheritableThreadLocal (引用)

    ThreadLocal:http://www.cnblogs.com/moonandstar08/p/4912673.html InheritableThreadLocal:  http://www. ...

  10. 【linux】 linux gpio操作

    欢迎转载,转载时需保留作者信息,谢谢. 邮箱:tangzhongp@163.com 博客园地址:http://www.cnblogs.com/embedded-tzp Csdn博客地址:http:// ...