HDU 1896

Stones

Time Limit: 5000/3000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 458 Accepted Submission(s): 264

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

#include <iostream>
#include <vector>
#include <queue>
using namespace std;

typedef pair<int,int> TWICE;

int main()
{
    int i,j,k,T; 
    int N, p, d;  
    cin>>T;
    while (T--) 
    {
        priority_queue <TWICE, vector <TWICE>, greater <TWICE> > q;//两个均排序 
        cin>>N;
        for (i = 1; i <= N; ++i) 
        {
            cin>>p>>d;
            q.push(make_pair(p,d));
        }
        int cnt = 0, max = 0;
        while (!q.empty()) 
        {
            cnt++;
            TWICE u = q.top(); 
            q.pop();
            if (cnt&1) 
               q.push(make_pair(u.first + u.second, u.second));
            else 
               max = u.first;
        }
        cout<<max<<endl;
    }   
    //while (1);
    return 0;
}

 运算符重载：

struct node         //直接在结构体里边定义小于运算符
{
     int position,length;
     bool operator < (const node &p) const     //定义小于操作符,也可以把&去掉，不用引用类型，时间一样
     {
         if(position != p.position)
             return position > p.position;
         else
             return length > p.length;
     }

};
priority_queue<node> s; //优先队列--从小到大排序