Regular polygon
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2219 Accepted Submission(s): 880
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
题目大意：有n个点，问总共能形成多少个正多边形。
思路：实际上只能形成正四边形，在官方题解里给出了相关的论文。。。然后就变成了一道水题，暴力枚举点对，查找每个点对是否有对应的点对形成正方形。由于坐标存在负数，对输入数据做一下处理，以及注意横（纵）坐标差值<=200，注意到这两点基本没问题了。具体看代码。
AC代码：
1 #include<iostream> 2 #include<cstring> 3 #include<cstdio> 4 #include<cmath> 5 using namespace std; 6 struct Node{ 7 int x; 8 int y; 9 }node[505]; 10 bool vis[601][601]; 11 int main() 12 { 13 int n,dx,dy,a,b; 14 while(~scanf("%d", &n)) 15 { 16 int res=0; 17 memset(vis, 0, sizeof(vis)); 18 for(int i=0;i<n;i++){ 19 scanf("%d%d", &a, &b); 20 a+=300; 21 b+=300; 22 node[i].x=a; 23 node[i].y=b; 24 vis[node[i].x][node[i].y]=1; 25 } 26 27 28 for(int i=0;i<n;i++) 29 for(int j=i+1;j<n;j++){ 30 dx=node[i].x-node[j].x; 31 dy=node[i].y-node[j].y; 32 if(vis[node[i].x+dy][node[i].y-dx]==1&&vis[node[j].x+dy][node[j].y-dx]==1) 33 res++; 34 if(vis[node[i].x-dy][node[i].y+dx]==1&&vis[node[j].x-dy][node[j].y+dx]==1) 35 res++; 36 37 } 38 cout<<res/4<<endl; 39 } 40 }