hdu 5437 Alisha’s Party
来源:互联网 发布:巨炮舰队扫矿软件 编辑:程序博客网 时间:2024/06/06 02:09
Alisha’s Party
Time Limit: 3000/2000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 652 Accepted Submission(s): 182
Problem Description
Princess Alisha invites her friends to come to her birthday party. Each of her friends will bring a gift of some value v , and all of them will come at a different time. Because the lobby is not large enough, Alisha can only let a few people in at a time. She decides to let the person whose gift has the highest value enter first.
Each time when Alisha opens the door, she can decide to letp people enter her castle. If there are less than p people in the lobby, then all of them would enter. And after all of her friends has arrived, Alisha will open the door again and this time every friend who has not entered yet would enter.
If there are two friends who bring gifts of the same value, then the one who comes first should enter first. Given a queryn Please tell Alisha who the n−th person to enter her castle is.
Each time when Alisha opens the door, she can decide to let
If there are two friends who bring gifts of the same value, then the one who comes first should enter first. Given a query
Input
The first line of the input gives the number of test cases, T , where 1≤T≤15 .
In each test case, the first line contains three numbersk,m and q separated by blanks. k is the number of her friends invited where 1≤k≤150,000 . The door would open m times before all Alisha’s friends arrive where 0≤m≤k . Alisha will have q queries where 1≤q≤100 .
Thei−th of the following k lines gives a string Bi , which consists of no more than 200 English characters, and an integer vi , 1≤vi≤108 , separated by a blank. Bi is the name of the i−th person coming to Alisha’s party and Bi brings a gift of value vi .
Each of the followingm lines contains two integers t(1≤t≤k) and p(0≤p≤k) separated by a blank. The door will open right after the t−th person arrives, and Alisha will let p friends enter her castle.
The last line of each test case will containq numbers n1,...,nq separated by a space, which means Alisha wants to know who are the n1−th,...,nq−th friends to enter her castle.
Note: there will be at most two test cases containingn>10000 .
In each test case, the first line contains three numbers
The
Each of the following
The last line of each test case will contain
Note: there will be at most two test cases containing
Output
For each test case, output the corresponding name of Alisha’s query, separated by a space.
Sample Input
15 2 3Sorey 3Rose 3Maltran 3Lailah 5Mikleo 61 14 21 2 3
Sample Output
Sorey Lailah Rose
注意:t非递减,有相同,处理完,优先队列搞一搞
#include <stdio.h>#include <algorithm>#include <string.h>#include <queue>int ans[150020];using namespace std;struct node{ int num; char name[201]; int v;}a[150020];struct node1{ int t; int p;}vv[150020];bool operator <(const node &a,const node &b){ if(a.v==b.v) { return a.num>b.num; } else return a.v<b.v;}bool cmp(node1 a,node1 b){ return a.t< b.t;}int main(){ int T; priority_queue<node>qu; scanf("%d", &T); while(T--) { int k, m, q; scanf("%d%d%d", &k, &m, &q); while(!qu.empty()) qu.pop(); for(int i = 1;i <= k;i++) { scanf("%s%d",a[i].name,&a[i].v); a[i].num = i; } for(int i = 1;i <= m;i++) { scanf("%d%d", &vv[i].t, &vv[i].p); } sort(vv+1,vv+m+1,cmp); for(int i = 1;i < m;i++) { if(vv[i].t==vv[i+1].t) { vv[i].p += vv[i+1].p; for(int j = i+1;j < m;j++) { vv[j] = vv[j+1]; } i--; m--; } } /*for(int pp = 1;pp<=m;pp++) { printf("%d %d\n",vv[pp].t,vv[pp].p); }*/ vv[0].t = 0; int now = 0; for(int i = 1;i <= m;i++) { for(int j =vv[i-1].t+1;j <= vv[i].t;j++) qu.push(a[j]); for(int l = 1;l <= vv[i].p;l++) { if(qu.empty()) break; ans[now+1] = qu.top().num; qu.pop(); now++; } } for(int j = vv[m].t+1;j <= k;j++) qu.push(a[j]); while(!qu.empty()) { ans[now+1] = qu.top().num; qu.pop(); now++; } int ques[102]; for(int i = 1;i <= q;i++) { scanf("%d", &ques[i]); } for(int j = 1;j <= q;j++) { if(j==1) printf("%s",a[ans[ques[j]]].name); else printf(" %s",a[ans[ques[j]]].name); } printf("\n"); }}
0 0
- hdu 5437 Alisha’s Party
- hdu 5437 Alisha’s Party
- HDU 5437 Alisha’s Party
- HDU 5437 Alisha’s Party
- HDU-5437 Alisha’s Party
- HDU 5437 Alisha’s Party
- HDU 5437 Alisha’s Party
- hdu 5437(Alisha’s Party)
- HDU 5437 Alisha’s Party
- HDU 5437 Alisha’s Party
- Alisha’s Party HDU
- HDU 5437 Alisha’s Party 优先队列
- hdu 5437 Alisha’s Party 优先队列
- HDU 5437 Alisha’s Party(模拟)
- hdu 5437Alisha’s Party(优先队列)
- hdu 5437 Alisha’s Party (优先队列)
- hdu 5437 Alisha’s Party 优先队列
- [HDU 5437]Alisha’s Party[模拟]
- 飘雪
- mantis与testlink实验流程
- Java 容器类深入整理笔记
- php 加密解密
- 2015.09.13
- hdu 5437 Alisha’s Party
- Java常用类练习3
- 【leetcode】Roman to Integer【java】
- 想考个数据库工程师证要做哪些
- 关于iOS和OS X废弃的API你需要知道的一切
- unity 显示帧率
- windows下 MySQL的表名大小写 移植到Linux 区分大小写
- cocos2d-x 3.2 物理小游戏教程2 block it 物理世界 墙壁
- 文章标题