Alisha’s Party Problem's Link Mean: Alisha过生日,有k个朋友来参加聚会,由于空间有限,Alisha每次开门只能让p个人进来,而且带的礼物价值越高就越先进入. 给出开门的操作:当第ti个人来时,打开门让pi个人进入. q个询问,对于每个询问qi,你需要回答第qi个进入房间的人的名字. analyse: 维护一个优先队列,模拟就行. 需要注意,由于priority_queue对重载运算符貌似会出错,所以使用multiset来做priority_queue使用…

题意:邀请k个朋友,每个朋友带有礼物价值不一,m次开门,每次开门让一定人数p(如果门外人数少于p,全都进去)进来,当最后所有人都到了还会再开一次门,让还没进来的人进来,每次都是礼物价值高的人先进.最后给出q个数,表示要输出第ni个进来的人的名字. 析:其实这就是一个模拟题,很容易知道是优先队列模拟,不能set,会超时,反正我是超时了,然后就一步步的模拟就好了,注意它给的时间可能不是按顺序,要排序,在比赛时我就没排序,一直WA.... 代码如下: #include <cstdio> #inclu…

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…

Alisha’s Party Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/contests/contest_showproblem.php?pid=1001&cid=621 Description Princess Alisha invites her friends to come to her birthday party. Each of her friends will bring a gift of…

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…

题意:有k个人带着价值vi的礼物来,开m次门,每次在有t个人来的时候开门放进来p个人,所有人都来了之后再开一次门把剩下的人都放进来,每次带礼物价值高的人先进,价值相同先来先进,q次询问,询问第n个进来的人的名字. 解法:一道现场wa到死的模拟……拿set/map/priority_queue维护一下序列.赛后重写一遍……依旧wa到死……我只想说…… m可以为0 m可以为0 m可以为0啊! orz…… 代码: #include<stdio.h> #include<iostream>…

Alisha’s Party Time Limit: 3000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 4389    Accepted Submission(s): 1121 Problem Description Princess Alisha invites her friends to come to her birthday party. Each…

Alisha 举办聚会,会在一定朋友到达时打开门,并允许相应数量的朋友进入,带的礼物价值大的先进,最后一个人到达之后放外面的所有人进来.用优先队列模拟即可.需要定义朋友结构体,存储每个人的到达顺序以及携带礼品价值,并重载<运算符.然后模拟即可. #include<iostream> #include<cstdio> #include<cstring> #include<vector> #include<algorithm> #include…

优先队列模拟一下就好. #include<bits/stdc++.h> using namespace std; priority_queue<int>q; int main() { int n; scanf("%d",&n); int t; scanf("%d",&t); ; i <= n; i++){ int v; scanf("%d",&v); q.push(v); } ; while(…

Alisha’s Party Time Limit: 3000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 7971    Accepted Submission(s): 1833 Description Princess Alisha invites her friends to come to her birthday party. Each of her f…