17多校contest two 1011Regular polygon ( 计算几何

最新推荐文章于 2020-03-15 21:38:33 发布

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online judge HDU 同时被 2 个专栏收录

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本文介绍了一种通过坐标判断平面上若干点能否构成正多边形的方法，并给出了一段AC代码实现。该方法主要利用数学公式推导，确定特定条件下正方形的存在可能性。

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Regular polygon

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

4
0 0
0 1
1 0
1 1
6
0 0
0 1
1 0
1 1
2 0
2 1

Sample Output

1
2

题意

平面上给一组坐标判断一共可以组成多少个正多边型

题解：

这个道题完全是靠结论的好不唉
坐标为整数的点所组成的正多边形一定是正方形
那么就是已知两个点枚举另外两个点（两个点可以决定2个正方形
已知(x1,x2) (y1,y2)

x3=x1+(y1-y2) y3= y1-(x1-x2)

x4=x2+(y1-y2) y4= y2-(x1-x2)

或

x3=x1-(y1-y2) y3= y1+(x1-x2)

x4=x2-(y1-y2) y4= y2+(x1-x2)

AC代码

#include <bits/stdc++.h>
using namespace std;

#define LL long long
#define CLR(a,b) memset(a,(b),sizeof(a))

const int N = 500+10;
bool vis[N<<1][N<<1];
int X[N], Y[N];

int main()
{
    int n;
    while(~scanf("%d",&n)) {
        CLR(vis,false);
        LL k = 0;
        for(int i = 0; i < n; i++) {
            scanf("%d%d",&X[i],&Y[i]);
            vis[N+X[i]][N+Y[i]] = true;
        }
        int x1, x2, x3,
         x4;
        int y1, y2, y3, y4;
        for(int i = 1; i < n; i++) {
            x1 = X[i], y1 = Y[i];
            for(int j = 0; j < i; j++) {
                x2 = X[j], y2 = Y[j];
                x3 = x1 + (y1-y2);
                y3 = y1 - (x1-x2);
                x4 = x2 + (y1-y2);
                y4 = y2 - (x1-x2);
                if(vis[N+x3][N+y3] && vis[N+x4][N+y4]) k++;
                x3 = x1 - (y1-y2);
                y3 = y1 + (x1-x2);
                x4 = x2 - (y1-y2);
                y4 = y2 + (x1-x2);
                if(vis[N+x3][N+y3] && vis[N+x4][N+y4]) k++;
            }
        }
        printf("%lld\n",k/4);
    }
return 0;
}