HDOJ--1896--Stones

题目描述：
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.
输入描述:
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.
输出描述:
Just output one line for one test case, as described in the Description.
输入：
2
2
1 5
2 4
2
1 5
6 6
输出：
11
12
题意：
路上有很多石头，当他遇到一块石头时，如果碰到的是第奇数个的石头，他会尽可能地把它扔到前面；如果碰到的是第偶数个的石头，他会把它留在原处。现在给您一些关于路上石头的信息，您要告诉我从起点到塞姆普路过的最远石头的距离。请注意，如果两个或更多的石头停留在同一位置，您将首先遇到较大的（具有最小Di的，如输入中所述）。
题解：
优先队列模拟一下
代码：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<queue>
using namespace std;

struct point {
    int pos;
    int dis;
    friend bool operator <(const point &a,const point &b) {
        if(a.pos == b.pos)
            return a.dis > b.dis;
        else
            return a.pos > b.pos;
    }
};

int main(){
    int t,n;
    scanf("%d",&t);
    while(t--){
        priority_queue <point> pq;
        point t;
        scanf("%d",&n);
        for(int i = 0; i < n; i ++){
            scanf("%d%d",&t.pos,&t.dis);
            pq.push(t);
        }
        int ans = 0;
        bool flag = true;
        point p;
        while(!pq.empty()){
            p = pq.top();
            pq.pop();
            if(flag){
                //ans = p.pos;
                p.pos += p.dis;
                pq.push(p);
            }
            flag = !flag;
        }
        printf("%d\n",p.pos);
    }
    return 0;
}