HDU 6055 组合正多边形问题

最新推荐文章于 2018-04-03 12:09:54 发布

原创最新推荐文章于 2018-04-03 12:09:54 发布 · 364 阅读

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Regular polygon

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 220 Accepted Submission(s): 86

Problem Description

On a two-dimensional plane, give you n integer points. Your task is to figure out how many different regular polygon these points can make.

Input

The input file consists of several test cases. Each case the first line is a numbers N (N <= 500). The next N lines ,each line contain two number Xi and Yi(-100 <= xi,yi <= 100), means the points’ position.(the data assures no two points share the same position.)

Output

For each case, output a number means how many different regular polygon these points can make.

Sample Input

Sample Output

1
2

Source

2017 Multi-University Training Contest - Team 2

题意:输入n个点，要求出这些点能组成多少个正多边形（边长内角全部相等）而且是整数点，那么既然是整数点，那么组成的图形只能是正方形

所以只要枚举两个点，旋转90度去找另外两个点是否存在，最后答案除4就可以了

#include<bits/stdc++.h>
using namespace std;
#define MAXN 800
struct point{
	int x,y;
}p[MAXN];
int ans;
int has[MAXN][MAXN];
point change(point a,point b){
	int x = a.x - b.x;
	int y = a.y - b.y;
	struct point c;
	c.x = b.x - y;
	c.y = b.y + x;
	return c;
}
bool isHas(point a){
	if(has[a.x][a.y])
		return true;
	return false;
} 
bool solve(int i,int j){
	point a = p[i];
	point b = p[j];
	point c = change(a,b);
	point d = change(b,c);
	if(isHas(c) && isHas(d))
		return true;
	return false;
}
int main(){
	int n;
	while(scanf("%d",&n) != EOF){
		memset(has,0,sizeof(has));
		for(int i = 1;i <= n;i++){
			scanf("%d %d",&p[i].x,&p[i].y);
			p[i].x += 300;
			p[i].y += 300;
			has[p[i].x][p[i].y] = 1;
		}
		ans = 0;
		for(int i = 1;i <= n;i++){
			for(int j = 1;j <= n;j++){
				if(i != j && solve(i,j)){
					ans++;
				}
			}
		}
		printf("%d\n",ans / 4);
	}
	return 0;
}