HDU 6055 - Regular polygon

Posted 2020-09-30 nicetomeetu

/*
HDU 6055 - Regular polygon [ 分析,枚举 ]
题意：
	给出 x,y 都在 [-100, +100] 范围内的 N 个整点，问组成的正多边形的数目是多少
	N <= 500
分析：
	分析可知，整点组成的正多边形只能是正方形
	故枚举两个点，验证剩下两个点的位置
	
	坑点： 由于点的范围是 [-100, +100]，故经过计算得出的点的范围可能是 [-300,+300]，注意越界

编码时长：46分钟(-1)
*/
#include <bits/stdc++.h>
using namespace std;
int n;
bool mp[1005][1005];
int ans;
int x[505], y[505];
void solve(int x1, int y1, int x2, int y2)
{
	if (x1 > x2) swap(x1, x2), swap(y1, y2);
	int x3, y3, x4, y4;
	x3 = x1 - (y2-y1);
	y3 = y1 + x2-x1;
	x4 = x2 - (y2-y1);
	y4 = y2 + x2-x1;
	if (mp[x3][y3] && mp[x4][y4]) ans++;
	x3 = x1 + y2-y1;
	y3 = y1 - (x2-x1);
	x4 = x2 + y2-y1;
	y4 = y2 - (x2-x1);
	
	if (mp[x3][y3] && mp[x4][y4]) ans++;
}
int main()
{
	while (~scanf("%d", &n))
	{
		memset(mp, 0, sizeof(mp));
		for (int i = 1; i <= n; i++)
		{
			scanf("%d%d", &x[i], &y[i]);
			x[i] += 500, y[i] += 500;
			mp[x[i]][y[i]] = 1;
		}
		ans = 0;
		for (int i = 1; i <= n; i++)
			for (int j = i+1; j <= n; j++)
				solve(x[i], y[i], x[j], y[j]);
		printf("%d\n", ans/4);
	}
}