hdu 1896 Stones(优先队列 Dijkstr)

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

11
12
题意：就是说给你一条自左向右的，路上会有石头，你遇到的第奇数个的石头，你要把它扔出去，第偶数个的石头不用管他，一行两个数，第一个代表石头的位置，第二个代
表这个石头能被扔多远，如果同一个地方有两块石头的话，先遇到扔得近的那块。
#include<iostream>#include<cstdio>#include<functional>#include<queue>#include<cstring>using namespace std;struct node {    int place;    int value;    friend bool operator < (node n1,node n2) {        if(n1.place!=n2.place)            return n1.place > n2.place;        else           return n1.value > n2.value;    }};int main() {    int t,n,m;    scanf("%d",&t);    while(t--){        scanf("%d",&n);        m = 0;        priority_queue<node>qn;        node d,D;        while(n--){         scanf("%d%d",&D.place,&D.value);         qn.push(D);        }        while(!qn.empty()){          D=qn.top();          m++;          qn.pop();          if(m&1){            d.place = D.place+D.value;            d.value = D.value;            qn.push(d);          }        }         printf("%d\n",D.place);    }}