V

V

Time Limit:3000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit Status

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

1112

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扔石头只扔第奇数个石头肉过两个石头在同一位置就扔远的那个

这也是优先队列的问题先优先排序位置然后排序大小

扔过的石头还有再次入列指导最后没有石头为止

123456789101112131415161718192021222324252627282930313233343536373839404142

#include<cstdio>#include<queue>struct node{int l;int f;bool friend operator < (node a,node b){if(a.l==b.l)return a.f>b.f;return a.l>b.l;}//按照优先排序的方法}a;using namespace std;int main(){priority_queue<node> cnt;int T;scanf("%d",&T);while(T--){int n;scanf("%d",&n);int i=1;for(i=1;i<=n;i++){scanf("%d%d",&a.l,&a.f);cnt.push(a);}int sum=0;i=1;while(!cnt.empty()){a=cnt.top();if(i&1){//判断这是石头是第几个石头sum=a.l+a.f;a.l=a.l+a.f;cnt.pop();cnt.push(a);}else{cnt.pop();}i++;}printf("%d\n",sum);}}