POJ - 2002：Squares

Squares

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题目

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.

输入

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.

输出

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

输入样例

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

输出样例

1
6
1

参考代码

#include
#include
#define MAXN 1000+5
using namespace std;
struct info{
	int x;
	int y;
};
bool cmp(info A, info B){
	if(A.x<B.x || A.x==B.x && A.y<B.y)
		return true;
	return false;
}
info pot[MAXN];
int n;

int cmp1(info A, info B){
	if(A.x<B.x || A.x==B.x && A.y<B.y)
		return -1;
	else if(A.x==B.x && A.y==B.y){
		return 0;
	}
	return 1;
}

int binSearch(info des){
	int start=0, end=n-1;
	int mid;
	while(start<=end){
		mid=(start+end)/2;
		if(cmp1(pot[mid],des)==-1){
			start=mid+1;
		}else if(cmp1(pot[mid],des)==1){
			end=mid-1;
		}else{
			break;
		}
	}
	if(pot[mid].x==des.x && pot[mid].y==des.y)
		return 1;
	return 0;
}
int main(){
	while(scanf("%d",&n)==1 && n){
		int i,j;
		for(i=0;i<n;i++){
			scanf("%d%d",&pot[i].x,&pot[i].y);
		}
		sort(pot,pot+n,cmp);
		double pos1x,pos1y,pos2x,pos2y;
		double midx,midy;
		int flag,cnt=0;
		for(i=0;i<n;i++){
			for(j=i+1;j<n;j++){
				flag=0;
				midx=(pot[i].x+pot[j].x)/2.0;
				midy=(pot[i].y+pot[j].y)/2.0;
				pos1x=midx+midy-pot[j].y;
				pos1y=midy-midx+pot[j].x;
				pos2x=midx+midy-pot[i].y;
				pos2y=midy-midx+pot[i].x;
				
				if(pos1x==int(pos1x) && pos1y==int(pos1y) && pos2x==int(pos2x) && pos2y==int(pos2y)){
					info pot1,pot2;
					pot1.x=pos1x;
					pot1.y=pos1y;
					pot2.x=pos2x;
					pot2.y=pos2y;
					flag+=binSearch(pot1)+binSearch(pot2);
					if(flag==2) cnt++;
				}
			}
		}
		printf("%d\n",cnt/2);
	}
	return 0;
}