poj 2002_Squares_ 4000 hash

题目描述

思路

#include <stdio.h>
#define M 999987
int h[M+1];
struct arr
{
int x,y;
};
arr a[M];
int insert(int x,int y)
{
int l=(x*50000+y);
int i=l%M;
if (l<0) i=-l%M;
while (h[i]!=-5201314&&h[i]!=l) i=i%M+1;
h[i]=l;
}
int find(int x,int y)
{
int l=(x*50000+y);
int i=l%M;
if (l<0) i=-l%M;
while (h[i]!=-5201314&&h[i]!=l) i=i%M+1;
if (h[i]==l) return true;
else return false;
}
int main()
{
int n;
scanf("%d",&n);
while (n!=0)
{
for (int i=1;i<=M;i++)
{
h[i]=-5201314;
}
for (int i=1;i<=n;i++)
{
scanf("%d%d",&a[i].x,&a[i].y);
insert(a[i].x,a[i].y);
}
int ans=0;

for (int i=1;i<=n;i++)
{
for (int j=i+1;j<=n;j++)
{
int x1,x2,y1,y2;
x1=a[i].x+(a[i].y-a[j].y);
y1=a[i].y-(a[i].x-a[j].x);
x2=a[j].x+(a[i].y-a[j].y);
y2=a[j].y-(a[i].x-a[j].x);
if (find(x1,y1)&&find(x2,y2)) ans++;
x1=a[i].x-(a[i].y-a[j].y);
y1=a[i].y+(a[i].x-a[j].x);
x2=a[j].x-(a[i].y-a[j].y);
y2=a[j].y+(a[i].x-a[j].x);
if (find(x1,y1)&&find(x2,y2)) ans++;
}
}
ans=ans>>2;
printf("%d\n",ans);
scanf("%d",&n);
}
}