HDUoj 1896 Stones ( 优先队列

Stones

Time Limit: 5000/3000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 2545    Accepted Submission(s): 1641

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

 

Author

Sempr|CrazyBird|hust07p43

很好的一个优先队列的问题

题意：路上有很多石头，当你遇到奇数序列的石头就把他向前仍，偶数的不动他，如果两个石头一起，先考虑可以仍的比较近的石头仍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仍了为止，求最远的石头距离起点多少用优先队列的最小堆写下就可以AC了

#include<cstdio>#include<queue>#include<algorithm>using namespace std;struct node{    int a, b;    bool friend operator < (node x,node y) {        if(x.a != y.a) {            return x.a > y.a;        }        return x.b > y.b;    }}p;int main(){    int T;    scanf("%d",&T);     while(T--) {        priority_queue<node> fq;        int n;        scanf("%d",&n);         for(int i = 0;i < n; i++) {            scanf("%d%d",&p.a,&p.b);            fq.push(p);        }        int k = 1;        while(!fq.empty()) {            p = fq.top();            fq.pop();            if(k%2) {                p.a += p.b;                fq.push(p);            }            k++;        }        printf("%d\n",p.a);    }return 0;}