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poj 2002 Square

2016-05-07 11:31 204 查看

Squares

Time Limit: 3500MS		Memory Limit: 65536K
Total Submissions: 18509		Accepted: 7131

Description

A square is a 4-sided polygon whose sides have equal length and adjacent sides form 90-degree angles. It is also a polygon such that rotating about its centre by 90 degrees gives the same polygon. It is not the only polygon with the latter property, however,
as a regular octagon also has this property.

So we all know what a square looks like, but can we find all possible squares that can be formed from a set of stars in a night sky? To make the problem easier, we will assume that the night sky is a 2-dimensional plane, and each star is specified by its x
and y coordinates.

Input

The input consists of a number of test cases. Each test case starts with the integer n (1 <= n <= 1000) indicating the number of points to follow. Each of the next n lines specify the x and y coordinates (two integers) of each point. You may assume that the
points are distinct and the magnitudes of the coordinates are less than 20000. The input is terminated when n = 0.
Output

For each test case, print on a line the number of squares one can form from the given stars.
Sample Input

Sample Output

1
6
1

Source

Rocky Mountain 2004

题意：
给定N个点，求出这些点一共可以构成多少个正方形。

思路：
先枚举2个点，之后算出另外两个点，判断是否符合

/*
由于我没有开个数组记录点的坐标，所以把cnt定为1005；
*/
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int M=20007;
struct node
{
int x,y,next;
}a[20007];
int head[20007],cnt;
bool cmp(node a,node b)
{
if(a.x==b.x) return a.y<b.y;
return a.x<b.x;
}
void insert(int x)//进行哈希并且用链地址法解决地址冲突
{
int h=(a[x].x*a[x].x+a[x].y*a[x].y)%M;
a[cnt].next=head[h];
a[cnt].x=a[x].x;
a[cnt].y=a[x].y;
head[h]=cnt++;
}
int find(int x,int y)
{
int h=(x*x+y*y)%M;
for(int i=head[h];i!=-1;i=a[i].next)
{
if(a[i].x==x&&a[i].y==y)
return i;
}
return -1;
}
int main()
{
int i,j,n;
while(scanf("%d",&n)&&n)
{
memset(head,-1,sizeof(head));
memset(a,0,sizeof(a));
cnt=1005;long long ans=0;
for(i=0;i<n;i++)
{
scanf("%d%d",&a[i].x,&a[i].y);
insert(i);
}
sort(a,a+n,cmp);
for(i=0;i<n;i++)
{
for(j=i+1;j<n;j++)
{
int x1=a[i].x-a[j].y+a[i].y;
int y1=a[i].y+a[j].x-a[i].x;
if(find(x1,y1)==-1) continue;
int x2=a[j].x-a[j].y+a[i].y;
int y2=a[j].y+a[j].x-a[i].x;
if(find(x2,y2)==-1) continue;
ans++;
}
}
printf("%d\n",ans/2);
}
return 0;
}

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