HDU

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1896点击打开链接

Stones

Time Limit: 5000/3000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 2885    Accepted Submission(s): 1845

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. 

 

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.

 

Output

Just output one line for one test case, as described in the Description.

 

Sample Input

221 52 421 56 6

 

Sample Output

1112

 

优先队列 距离近的先出来判断

注意奇数偶数不同情况

#include <iostream>#include <queue>#include <stdio.h>#include <stdlib.h>#include <stack>#include <limits>#include <string>#include <string.h>#include <vector>#include <set>#include <map>#include <algorithm>#include <math.h>using namespace std;struct xjy{    int now;    int step;    bool operator < ( const xjy &r)const    {        if(now==r.now)            return step>r.step;        return now>r.now;    }};priority_queue<xjy>q;int main(){    int t=0;    scanf("%d",&t);    while(t--)    {        while(!q.empty())            q.pop();        int n=0;        scanf("%d",&n);        for(int i=0;i<n;i++)            {                xjy mid;                scanf("%d%d",&mid.now,&mid.step);                q.push(mid);            }        int num=1;        int ans=0;        while(!q.empty())        {            xjy mid;            mid=q.top();            q.pop();            if(num&1)            {                ans=max(ans,mid.now);                mid.now+=mid.step;                q.push(mid);                num++;            }            else            {                ans=max(ans,mid.now);                num++;            }        }        printf("%d\n",ans);    }}