Stones(优先队列)
来源:互联网 发布:人工智能的编程语言 编辑:程序博客网 时间:2024/05/21 09:02
Stones
Time Limit: 5000/3000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)Total Submission(s): 1272 Accepted Submission(s): 784
Problem Description
Because of the wrong status of the bicycle, Sempr begin to walk east to west every morning and walk back every evening. Walking may cause a little tired, so Sempr always play some games this time.
There are many stones on the road, when he meet a stone, he will throw it ahead as far as possible if it is the odd stone he meet, or leave it where it was if it is the even stone. Now give you some informations about the stones on the road, you are to tell me the distance from the start point to the farthest stone after Sempr walk by. Please pay attention that if two or more stones stay at the same position, you will meet the larger one(the one with the smallest Di, as described in the Input) first.
There are many stones on the road, when he meet a stone, he will throw it ahead as far as possible if it is the odd stone he meet, or leave it where it was if it is the even stone. Now give you some informations about the stones on the road, you are to tell me the distance from the start point to the farthest stone after Sempr walk by. Please pay attention that if two or more stones stay at the same position, you will meet the larger one(the one with the smallest Di, as described in the Input) first.
Input
In the first line, there is an Integer T(1<=T<=10), which means the test cases in the input file. Then followed by T test cases.
For each test case, I will give you an Integer N(0<N<=100,000) in the first line, which means the number of stones on the road. Then followed by N lines and there are two integers Pi(0<=Pi<=100,000) and Di(0<=Di<=1,000) in the line, which means the position of the i-th stone and how far Sempr can throw it.
For each test case, I will give you an Integer N(0<N<=100,000) in the first line, which means the number of stones on the road. Then followed by N lines and there are two integers Pi(0<=Pi<=100,000) and Di(0<=Di<=1,000) in the line, which means the position of the i-th stone and how far Sempr can throw it.
Output
Just output one line for one test case, as described in the Description.
Sample Input
221 52 421 56 6
Sample Output
1112
Author
Sempr|CrazyBird|hust07p43
Source
HDU 2008-4 Programming Contest
是奇数就扔,是偶数就忽略!记住就是在同一点的石头不算一个
有几个算几个!优先队列!
关于优先队列的详解看这里!优先队列
#include<cstdio>#include<iostream>#include<queue>#include<vector>using namespace std;struct node{ int pos; int dis;};struct cmp{ bool operator () (node a,node b){ if(a.pos!=b.pos) return a.pos>b.pos; else return a.dis>b.dis; }};int main(){ priority_queue<node,vector<node>,cmp>s; int t,n1,num,result; cin>>t; while(t--) { node n; cin>>n1; while(n1--) { cin>>n.pos>>n.dis; s.push(n); } num=1; while(!s.empty()) { n=s.top(); s.pop(); if(num&1) { n.pos+=n.dis; s.push(n); } num++; result=n.pos; } cout<<result<<endl; }}
struct cmp{
bool operator () (node a,node b){
if(a.pos!=b.pos)
return a.pos>b.pos;
else
return a.dis>b.dis;
}
};
这里要说一下,为什么这个明明排序是头大尾小,为什么输出是
头小尾大呢?
看这个吧,可能与这个机制有关记住就行了!
0 0
- Stones(优先队列)
- Stones(优先队列)
- Stones(hdu1896)优先队列
- 1896 Stones(优先队列)
- HDOJ Stones (优先队列)
- hd1896 Stones(队列优先)
- HDU Stones(优先队列)
- HDU 1896 -- Stones (优先队列)
- 杭电1896 Stones(优先队列)
- hdu1896 Stones(优先队列水)
- HDU 1896 Stones(优先队列)
- HDOJ 1896 Stones(优先队列)
- 【HDU]-1896-Stones(优先队列,好)
- HDU:1896 Stones(优先队列)
- 杭电-1896 Stones(优先队列)
- HDU 1896 Stones (优先队列)
- hdu--1896Stones(优先队列)
- HDU 1896:Stones(优先队列)
- mac下svn管理工具 Versions,一启动就crash的解决办法
- [Usaco-3.2.6] Sweet Butter香甜的黄油
- 关于pdo为何自动转换类型为string的问题
- .NET泛型解析(下)
- 树的基本运用一
- Stones(优先队列)
- Java数组实现循环队列的两种方法
- 删除Xcode中多余的证书provisioning profile
- 修改 QQ 聊天记录保存路径 ,并禁止随意更改
- 工作周报047
- 优先队列详解
- python yield keyword
- 批处理命令学习
- Oracle数据库之表空间