hdu 2428（哈希）

Stars

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)

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
解题思路：这道题目就是找到两点，然后查看另外两点是否存在。。。。典型的哈希
AC:
#include<iostream>#include<cstdio>#include<cstring>using namespace std;const int maxn = 1010;struct node{int x,y;}star[maxn];struct link{int id;int next;}Hash[maxn];int n,h[maxn*3],cnt;int calc(int x,int y){int key = x*x + y*y;return key % 1003;}void add(int id){int key = calc(star[id].x,star[id].y);Hash[cnt].id = id;Hash[cnt].next = h[key];h[key] = cnt++;}bool find(int x, int y){int key = calc(x,y);int p = h[key];while(p != -1){int t = Hash[p].id;if(star[t].x == x && star[t].y == y)return true;p = Hash[p].next;}return false;}int main(){int t;scanf("%d",&t);while(t--){scanf("%d",&n);memset(h,-1,sizeof(h));memset(Hash,-1,sizeof(Hash));for(int i = 1; i <= n; i++){scanf("%d%d",&star[i].x,&star[i].y);add(i);}int ans = 0;for(int i = 1; i <= n; i++)for(int j = i+1; j <= n; j++){if(star[i].x == star[j].x)//横坐标相等，往左右两面找正方形{int dy = abs(star[i].y - star[j].y);if(find(star[i].x + dy, star[i].y) && find(star[j].x + dy, star[j].y))ans++;if(find(star[i].x - dy, star[i].y) && find(star[j].x - dy, star[j].y))ans++;}else if(star[i].y == star[j].y) //纵坐标相等，往上下找正方形{int dx = abs(star[i].x - star[j].x);if(find(star[i].x, star[i].y + dx) && find(star[j].x, star[j].y + dx))ans++;if(find(star[i].x, star[i].y - dx) && find(star[j].x, star[j].y - dx))ans++;}}printf("%d\n",ans>>2);}return 0;}