hdu2428 Stars

Stars

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 1764    Accepted Submission(s): 612

Problem Description

    Lucy loves stars very much. There are N (1 <= N <= 1000) stars in the sky. Assume the sky is a flat plane. All of the stars lie on it with a location (x, y), -10000 <= x, y <= 10000. Now, Lucy wants you to tell her how many squares with each of four vertices formed by these stars, and the edges of the squares should parallel to the coordinate axes.

 

Input

    The first line of input is the number of test case.
    The first line of each test case contains an integer N. The following N lines each contains two integers x, y, meaning the coordinates of the stars. All the coordinates are different.

 

Output

    For each test case, output the number of squares in a single line.

 

Sample Input

211 240 01 01 10 1

 

Sample Output

01

 

全哈希。

#include<cstdio>#include<cstring>#include<string.h>using namespace std;const int maxn=1005;const int hashsize=2006;struct Point{int x;int y;Point(int a=0,int b=0){x=a;y=b;}}stars[maxn];int head[hashsize];int pre[maxn];int hash(Point p){int c=hashsize;return ((p.x*20005+p.y)%c+c)%c;}int find(int x,int y){int f=hash(Point(x,y));for(int i=head[f];i>=0;i=pre[i]){if(stars[i].x==x&&stars[i].y==y)return 1;}return 0;}int main(){int casen;scanf("%d",&casen);while(casen--){int n;scanf("%d",&n);memset(head,-1,sizeof(head));for(int i=0;i<n;i++){scanf("%d%d",&stars[i].x,&stars[i].y);int f=hash(stars[i]);pre[i]=head[f];head[f]=i;}int sum=0;for(int i=0;i<n;i++){for(int j=0;j<n;j++){if(stars[i].x>=stars[j].x)continue;if(stars[i].x-stars[j].x!=stars[i].y-stars[j].y)continue;if(!find(stars[i].x,stars[j].y))continue;if(!find(stars[j].x,stars[i].y))continue;sum++;}}printf("%d\n",sum);}return 0;}