Regular polygon HDU - 6055

最新推荐文章于 2017-08-21 20:53:04 发布

原创最新推荐文章于 2017-08-21 20:53:04 发布 · 292 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#计算几何 #acm

计算几何专栏收录该内容

7 篇文章

订阅专栏

本文介绍了一种算法，该算法通过输入一系列整数坐标点来找出这些点能构成的不同正方形的数量。通过枚举点对并检查是否存在对应的对角点对来实现。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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

一道简单的几何题，由于输入的是整数点，那么在整数点上只能找到正四边形，其余的都是无法构造的。所以枚举输入的任意两点，以这两点连成的线段为对角线，看另外两个点是否存在就可以得到答案了。

#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <cstring>
#include <math.h>
using namespace std;
int n;
struct point{
    int x, y;
};
point v[510];
int map1[500][500];
bool check(int x)
{
    if (-100 <= x && x <= 100)
        return true;
    else return false;
}
int main()
{
    while(scanf("%d", &n) != EOF) {
        memset(map1, 0, sizeof(map1));
        for (int i = 1; i <= n; i++) {
            scanf("%d%d", &v[i].x, &v[i].y);
            map1[v[i].x+100][v[i].y+100] = 1;
        }
        long long int sum = 0;
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= n; j++) if (i != j){
                int x1 = v[j].x + v[i].x;
                int y1 = v[j].y + v[i].y;
                int x2 = (v[j].y - v[i].y + x1) ;
                int y2 = (v[i].x - v[j].x + y1) ;
                if (x2 % 2 != 0 || y2 % 2 != 0) continue;
                    x2 /= 2;
                    y2 /= 2;
                if (check(x2) && check(y2) && map1[x2+100][y2+100]) {
                    int x3 = (v[i].y - v[j].y + x1);
                    int y3 = (v[j].x - v[i].x + y1);
                    if (x3 % 2 != 0 || y3 % 2 != 0) continue;
                    x3 /= 2;
                    y3 /= 2;
                    if (check(x3) && check(y3) && map1[x3+100][y3+100])
                        sum++;
                }
            }
        }
        cout <<sum/4<<endl;
    }
    return 0;
}