HDU 4007 Dave

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Dave

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65768/65768 K (Java/Others)
Total Submission(s): 3989 Accepted Submission(s): 1343

Problem Description

Recently, Dave is boring, so he often walks around. He finds that some places are too crowded, for example, the ground. He couldn't help to think of the disasters happening recently. Crowded place is not safe. He knows there are N (1<=N<=1000) people on the ground. Now he wants to know how many people will be in a square with the length of R (1<=R<=1000000000). (Including boundary).

Input

The input contains several cases. For each case there are two positive integers N and R, and then N lines follow. Each gives the (x, y) (1<=x, y<=1000000000) coordinates of people.

Output

Output the largest number of people in a square with the length of R.

Sample Input

Sample Output

  
   3


   
    
     Hint
    If two people stand in one place, they are embracing.

Source

The 36th ACM/ICPC Asia Regional Dalian Site —— Online Contest

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题目大意：

给出 n 个坐标给出一个边长为 r 的正方形然后问最多有多少个点会在这个正方形中

思路：

找出所有点中的最高的和最低的，然后开始枚举枚举x在里面的然后取最值

AC代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
#include<cmath>
#include<stdlib.h>
#include<map>
using namespace std;
#define inf 1<<30
#define N 1006
int n,r;
struct Node{
    int x,y;
}p[N];
int yy[N];
int xx[N];
int main()
{
    while(scanf("%d%d",&n,&r)==2)
        {
            for(int i=0;i<n;i++)
            {
                scanf("%d%d",&p[i].x,&p[i].y);
                yy[i]=p[i].y;
            }
            sort(yy,yy+n);

            int ans=0;
            for(int i=0;i<n;i++)
            {
                int xcnt=0;
                for(int j=0;j<n;j++)
                {
                    if(p[j].y<=yy[i]+r && p[j].y>=yy[i])
                    {
                        xx[xcnt++]=p[j].x;
                    }
                }
                sort(xx,xx+xcnt);
                int e=0;
                for(int j=0;j<xcnt;j++){
                    while(xx[e]<=xx[j]+r && e<xcnt) e++;
                    ans=max(ans,e-j);
            }

        }
        printf("%d\n",ans);
    }
    return 0;
}