[HDU]-6055 Regular polygon

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本篇博客介绍了一道算法题目，该题目要求在二维平面上通过给定的整数点集找出能构成的不同正多边形的数量。文章提供了一个具体的解题思路和实现代码，采用枚举两个点来确定其它点的方法，并通过判断点的位置来统计正多边形数目。

Regular polygon

URL: http://acm.hdu.edu.cn/showproblem.php?pid=6055

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 86 Accepted Submission(s): 31

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
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

题解

其实知道两个点，另外两个点就确定了，枚举两个点即可 $o(n^2)$

#include<stdio.h>
#include<string.h>
const int MAXN = 250;
int num[MAXN][MAXN],n;
struct node{int x, y;} dat[650];

int main()
{
    while(scanf("%d", &n)!=EOF){
        memset(num, 0 ,sizeof(num));
        for(int i = 0; i < n; ++i) {
            int X, Y;
            scanf("%d%d", &X, &Y);
            X += 110;
            Y += 110;
            ++num[X][Y];
            dat[i].x = X;
            dat[i].y = Y;
        }
        int ans = 0;
        for(int i = 0; i < n; ++i){
            for(int j = i + 1; j < n; ++j){
                int t1 = 0, t2 = 0;
                int x1 = dat[j].x - dat[j].y + dat[i].y;
                int y1 = dat[j].y + dat[j].x - dat[i].x;
                int x2 = dat[j].x - dat[j].y + dat[i].y - dat[j].x + dat[i].x;
                int y2 = dat[j].y + dat[j].x - dat[i].x - dat[j].y + dat[i].y;

                if(0 < x1 && x1 < MAXN && 0 < y1 && y1 < MAXN && 0 < x2 && x2 < MAXN && 0 < y2 && y2 < MAXN)
                    t1 = num[dat[i].x][dat[i].y] && num[dat[j].x][dat[j].y] && num[x1][y1] && num[x2][y2];

                x1 = dat[j].x + dat[j].y - dat[i].y;
                y1 = dat[j].y - dat[j].x + dat[i].x;
                x2 = dat[j].x + dat[j].y - dat[i].y - dat[j].x + dat[i].x;
                y2 = dat[j].y - dat[j].x + dat[i].x - dat[j].y + dat[i].y;
                if(0 < x1 && x1 < MAXN && 0 < y1 && y1 < MAXN && 0 < x2 && x2 < MAXN && 0 < y2 && y2 < MAXN)
                    t2 = num[dat[i].x][dat[i].y] && num[dat[j].x][dat[j].y] && num[x1][y1] && num[x2][y2];

                if(t1) ++ans;
                if(t2) ++ans;

             }
        }
        printf("%d\n",ans/4);
    }
    return 0;
}