Stones

Time Limit: 5000/3000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)

Total Submission(s): 1474    Accepted Submission(s): 921

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

/*题目大意是路上有很多石头，当你遇到奇数序列的石头就把他向前仍，偶数的不动，两个石头一起，先考虑可以扔的比较近的石头也就是比较大的石头，这样一直下去，直到前面所有的石头都不可以扔了为止，求最远的石头距离起点多少！！
题目这题用优先队列非常方便
*/
#include<stdio.h>
#include<queue>
#include<iostream>
using namespace std;
struct line//定义一个结构体，分别存储石头现在的位置和能扔出去的距离到优先队列中！
{
	int x;
	int dis;
	friend bool operator<(line a,line b)//然后每次取“最小的”！ 要特别注意多个石头的x一样的情况,要优先考虑y值最小的那块石头
	{
		if(a.x ==b.x )
		return a.dis >b.dis ;
		return a.x >b.x ;
	}
};
int main()
{
	int m,n;
	priority_queue<line > Q;
	scanf("%d",&n);
	line temp;
	while(n--)
	{
		scanf("%d",&m);
	    while(m--)
		{
			scanf("%d%d",&temp.x ,&temp.dis );
		     Q.push(temp);
		}
		int sum=1;
		while(!Q.empty() )//然后每次取“最小的”，如果取得是偶数个就不动，取得是奇数个就要更新该石头的位置并重新存到优先队列中，直到队列空，输出最后一个石头的位置
		{
		temp=Q.top() ;
		Q.pop() ;
		if(sum%2)
		{
			temp.x =temp.x +temp.dis ;
			Q.push(temp); 

		}sum++;
		}
		printf("%d\n",temp.x );
	}
	return 0;
} 

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