Regular polygon（HDU 6055）

Regular polygon

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 138    Accepted Submission(s): 52

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

40 00 11 01 160 00 11 01 12 02 1

 

Sample Output

12

 

Source

2017 Multi-University Training Contest - Team 2

 //题意：二维平面上给N个整数点，问能构成几个不同的正多边形。

//思路：这种正多边形只能是正四边形，即正方形（因为给的位置点的坐标都是整数，其他多边形不可能每条边都相同，好像有论文证明的QAQ）。这样问题就简化了很多。

把这几种情况考虑全就行，注意菱形不是正4变形...因为这个WA了无数发...，这题应该是team2最水的题之一了，然而WA了20发...20发...

还有一点就是坐标是可能为负的（但>= -100），所以用 map[][]去存的时候要点的坐标+105，不然map[][]这2个[][]里的数字会是负的，会RE...

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <cmath>#include <queue>#include <algorithm>using namespace std;const int MAX = 600;const int INF = -1e9;typedef struct {int x, y;}Point;int n;long long sum;int map[3000][3000];int maxx, maxy;Point point[MAX];bool cmp(Point a, Point b){if (a.y == b.y)return a.x < b.x;return a.y < b.y;}void find(){int i, j, k;for (i = 0; i < n; i++){if (point[i].x == maxx || point[i].y == maxy)continue;for (j = i + 1; j < n; j++){if (point[j].x <= point[i].x)continue;if (point[j].y <= point[i].y)continue;int len = point[j].x - point[i].x;int leny = point[j].y - point[i].y;if (map[point[i].x + len + 105][point[i].y + leny + 105] == 1 && map[point[i].x + leny + 105][point[i].y - len + 105] == 1 && map[point[i].x + len + leny + 105][point[i].y + leny - len + 105] == 1){sum++;}if (point[j].y == point[i].y + len){if (map[point[i].x + len + 105][point[i].y + 105] == 1 && map[point[i].x + 105][point[i].y + len + 105] == 1 && map[point[i].x + len + 105][point[i].y + len + 105] == 1){sum++;}}}}}int main(){while (scanf("%d", &n) != EOF){sum = 0;maxx = INF;maxy = INF;memset(map, 0, sizeof(map));for (int i = 0; i < n; i++){scanf("%d%d", &point[i].x, &point[i].y);int x = point[i].x + 105;int y = point[i].y + 105;map[x][y] = 1;if (point[i].x > maxx)maxx = point[i].x;if (point[i].y > maxy)maxy = point[i].y;}sort(point, point + n, cmp);find();printf("%lld\n", sum);}return 0;}