HDU 1896 Stones（优先队列）

Stones

Time Limit: 5000/3000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 1704    Accepted Submission(s): 1099

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

221 52 421 56 6

 

Sample Output

1112

 

Author

Sempr|CrazyBird|hust07p43

 

Source

HDU 2008-4 Programming Contest

 

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题意：给出石头的位置，和石头可以被扔出多远的距离，如果是偶数个石头就留在原地，如果是奇数个石头就扔出，如果石头再相同位置就扔出大的，所以要利用优先队列

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