【HDU 杭电 1896 】Stones

Stones

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
2
2
1 5
2 4
2
1 5
6 6

Sample Output
11
12

题意：sempr走在路上会遇见一些石头，如果遇见第奇数个石子，则将它扔到前面，如果是偶数，则什么也不做，如果某一个位置上有多个石头，则先遇见扔的比较近的那个，现在给出一些石头的初始位置和能够扔的距离，问到最后最远处的石头离初始位置多远！

思路：用优先队列，自定义优先级，当位置小的优先级最大，位置相同时D小的优先级较大,注意的是扔过石头也要加入队列，所有这道题适合用队列～～～

#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
struct node{
    int dis,pos;
    friend bool operator <(node a, node b)
    {
        if(a.pos==b.pos)
        return a.dis>b.dis;
        return a.pos>b.pos;
    }
}st;
int main()
{
    int t,i,n;
    scanf("%d",&t);
    priority_queue <node > q;
    while(t--)
    {
        scanf("%d",&n);
        while(!q.empty()) 
        q.pop();
        for(i=0;i<n;i++)
        {
            scanf("%d%d",&st.pos,&st.dis);
            q.push(st);
        }
        int ans=0,count=1;
        while(!q.empty())
        {
            if(count&1)
            {
                st=q.top();
                q.pop();
                st.pos+=st.dis;
                ans+=st.pos;
                q.push(st);
            }
            else
            q.pop();
            count++;
        }
        printf("%d\n",q.top().pos);
    }
    return 0;
}