Squares
Time Limit:3500MS |
|
Memory Limit:65536K |
Total Submissions:11102 |
|
Accepted:4021 |
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
4
1 0
0 1
1 1
0 0
9
0 0
1 0
2 0
0 2
1 2
2 2
0 1
1 1
2 1
4
-2 5
3 7
0 0
5 2
0
Sample Output
1
6
1
#include<iostream>
#include<cmath>
int n,i,j,ans;
using namespace std;
struct set
{
int x,y;
};
set point[1005];
set c,d;
int cmp(const void *a,const void *b)
{
if(((set *)a)->x==((set *)b)->x)
return ((set *)a)->y-((set *)b)->y;
return ((set *)a)->x-((set *)b)->x;
}
int fab(int a)
{
return a>0?a:-a;
}
bool search(set p,int n)
{
int max=n,min=1,mid;
while(max>=min)
{
mid=(max+min)/2;
if(cmp(&point[mid],&p)>0)
{
max=mid-1;
}
else if(cmp(&point[mid],&p)<0)
{
min=mid+1;
}
else
return true;
}
return false;
}
int main()
{
while(~scanf("%d",&n))
{
ans=0;
if(n==0)
break;
for(i=1;i<=n;i++)
scanf("%d%d",&point[i].x,&point[i].y);
qsort(point+1,n,sizeof(set),cmp);
for(i=1;i<=n;i++)
{
for(j=i+1;j<=n;j++)
{
int sign=(point[i].x-point[j].x)*(point[i].y-point[j].y);
c.x=point[i].x+fab(point[j].y-point[i].y);
c.y=point[i].y+fab(point[j].x-point[i].x);
d.x=point[j].x+fab(point[j].y-point[i].y);
d.y=point[j].y+fab(point[j].x-point[i].x);
if(sign<0)
{
if(search(c,n) && search(d,n))
ans++;
}
else if(sign==0 && point[i].x==point[j].x)
{
if(search(c,n) && search(d,n))
ans++;
}
}
}
printf("%d\n",ans);
}
return 0;
}
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