Hdu 1896 Stones(プライオリティキュー)

Stones
Time Limit : 5000/3000ms (Java/Other)   Memory Limit : 65535/32768K (Java/Other)
Total Submission(s) : 41   Accepted Submission(s) : 34
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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
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

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标题:1本の直線道路にn個の石があるということで、前へ進むと、1つの数に遭遇し、奇数番目に遭遇した場合はこの石を前へ距離D[i]を投げ、偶数番目であれば放置しておく.遭遇した最後の石の距離の出発点の位置を聞く.やってみると奇数番目の石に遭遇するたびに、それをD[i]に加えるのが簡単であるあ、優先列に戻して、石をもう一つ抜きます

#include
#include
#include
#include
using namespace std;

struct node
 {
     int x, t;
 };
 
 struct cmp
 {
     bool operator () (const node a, const node b)
    {
         if(a.x != b.x)return a.x > b.x;
        else return a.t > b.t;
     }
};


    

int main()
{
    int o,n;
priority_queue,cmp>q;
node a[100005],k;
scanf("%d",&o);
while(o--)
{
    while(!q.empty())
        q.pop();
scanf("%d",&n);
for(int i=0;i