hdu 2492 树状数组 Ping pong

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Ping pong

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4589    Accepted Submission(s): 1681

Problem Description

N(3<=N<=20000) ping pong players live along a west-east street(consider the street as a line segment).

Each
player has a unique skill rank. To improve their skill rank, they often
compete with each other. If two players want to compete, they must
choose a referee among other ping pong players and hold the game in the
referee's house. For some reason, the contestants can’t choose a referee
whose skill rank is higher or lower than both of theirs.

The
contestants have to walk to the referee’s house, and because they are
lazy, they want to make their total walking distance no more than the
distance between their houses. Of course all players live in different
houses and the position of their houses are all different. If the
referee or any of the two contestants is different, we call two games
different. Now is the problem: how many different games can be held in
this ping pong street?

 

Input

The
first line of the input contains an integer T(1<=T<=20),
indicating the number of test cases, followed by T lines each of which
describes a test case.

Every test case consists of N + 1
integers. The first integer is N, the number of players. Then N distinct
integers a1, a2 … aN follow, indicating the skill rank of each player,
in the order of west to east. (1 <= ai <= 100000, i = 1 … N).

 

Output

For each test case, output a single line contains an integer, the total number of different games.

 

Sample Input

1
3 1 2 3

 

Sample Output

1

 

Source

2008 Asia Regional Beijing

 

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#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
struct node{
    int u,v,m,h;
}que[];
bool cmp(struct node t1,struct node t2){
     if(t1.m==t2.m)
        return t1.u<t2.u;
     else
        return t1.m<t2.m;
}
int main(){
    int n;
    while(scanf("%d",&n)!=EOF){
            if(n==)
            break;
    for(int i=;i<n;i++){
        scanf("%d%d",&que[i].u,&que[i].v);
        que[i].h=(que[i].v-que[i].u)/+;
        que[i].m=que[i].u+que[i].h;
    }
    sort(que,que+n,cmp);
    int flag=;
    for(int i=;i<n-;i++){
            if(que[i].m>=que[i+].m){
                flag=;
                break;
            }
        if(que[i].m<que[i+].u)
            continue;
        double temp=que[i].m-que[i+].u;
        que[i+].m=temp+que[i+].h+que[i+].u;
    }
    if(flag)
        printf("NO\n");
    else
        printf("YES\n");

    }
   return ;
}