HDOJ 5071 Chat 模拟
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

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.
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.
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
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: 2HintThis problem description does not relate to any real person in THU.
#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
- 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 ...
- hdu 5071 Chat(模拟|Splay)
Chat Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others) Total Sub ...
- 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 ...
- 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 ...
- 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 ...
- 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 ...
- 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, ...
- 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 ...
- 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
- BZOJ 2748 音量调节
这道题我开始做时想用搜索来做,但是失败了,后来仔细一想发现这就是一个背包问题,之后一切就简单多了. 代码如下: #include<cstdio> #includ ...
- Median of Two Sorted Arrays-分治法
题目意思很简单将两个有序数组合并之后的中位数找出来.题目要求使用log(m+n)的时间复杂度来做. 虽然言简意赅,但不得不承认这个题目我自己想了好久也没做出来,隐约觉得应该使用寻找第k大数的算法来做, ...
- C语言新学备忘_1
#include <stdio.h> //C语言的标准输入 ,输出头文件扩展名为.h的文件称为头文件 //include称为文件包含命令 #include <stdlib.h> ...
- QT模态对话框用法(在UI文件中设置Widget背景图,这个图是一个带阴影边框的图片——酷)
QT弹出模态对话框做法: 1.新建UI文件时,一定要选择基类是QDialog的,我的选择是:Dialog without Buttons(),如下图: 2.然后在使用的时候: MyDialog dlg ...
- Runtime.getRuntime().exec(...)使用方法
Runtime.getRuntime().exec(...)使用方法 如果想要了解更多的信息,参阅代码里面给的链接 下面是这个正确的例子 public class RuntimeExec { /** ...
- Exam(贪心)
Exam Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submis ...
- EntityFramework经典的left join语法
/* * 常常看到有人问linq语法怎样写left join的查询语句,但网上找到的都是简单的两表连接.參考意义有限. * 今天最终项目里要用到复杂的多表连接,同一时候含有多个左连接, * 恰好 ...
- android -- 蓝牙 bluetooth (三)搜索蓝牙
接上篇打开蓝牙继续,来一起看下蓝牙搜索的流程,触发蓝牙搜索的条件形式上有两种,一是在蓝牙设置界面开启蓝牙会直接开始搜索,另一个是先打开蓝牙开关在进入蓝牙设置界面也会触发搜索,也可能还有其它触发方式,但 ...
- STL之vector详解
一.vector容器的自增长 首先,我们知道vector容器是由数组做出来的:它具备了数组的优缺点. 数组的优点: 操作数据,读取速度很快,因为有下标: 数组的缺点: 分配之后不能在改变大小: #in ...
- 再见了acm
2013年11月17日长沙区域赛我的最后一场区域赛. 忙碌了三年的acm要停下脚步,一时还无法接受. 这样一个结果有点无奈. 感谢队友,三年三支队伍五个队友,感谢你们.(每当写到这里时就总有点小忍不住 ...