Regular polygon
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1497 Accepted Submission(s): 582
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
Source
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题目大意：
读入n个点的坐标(x,y)，问能组成多少正方形
题目中说找正多边形，正多边形的要求：各条边相等，各个角相等。然而要在格点上，只有正方形才符合要求。
题解：
已知正方形对角两点坐标（a，b） 与 （c，d），经推导，则另外对角两点坐标为( ( a + b + c - d ) / 2 ，( - a + b + c + d ) / 2 ) 与 ( ( a - b + c + d ) / 2 , ( a + b - c + d ) / 2 )
注意下标越界问题，和与原来的对角的两个点重合。
#include <iostream> //#include<bits/stdc++.h> #include<cstdio> #include<algorithm> #include<cstring> #include<cmath> using namespace std; struct node { int x,y; }a[505]; int n,sum; bool vis[205][205]; bool ok( int x,int y) //是否越过界线 { if (x<0 || x>200 || y<0 || y>200) return 0; return 1; } bool same(int x,int y,int p,int q) //判断和原来对角线的两个点有没有重合 { if (x==a[p].x && y==a[p].y) return 0; if (x==a[q].x && y==a[q].y) return 0; return 1; } int main() { while(~scanf("%d",&n)) { memset(vis,0,sizeof(vis)); for(int i=1;i<=n;i++) { int x,y; scanf("%d%d",&x,&y); a[i].x=x+100; a[i].y=y+100; vis[a[i].x][a[i].y]=1; } sum=0; for(int i=1;i<=n;i++) for(int j=i+1;j<=n;j++) { int x3=(a[i].x+a[i].y+a[j].x-a[j].y); int y3=(-a[i].x+a[i].y+a[j].x+a[j].y); int x4=(a[i].x-a[i].y+a[j].x+a[j].y); int y4=(a[i].x+a[i].y-a[j].x+a[j].y); if (x3%2!=0 || y3%2!=0 || x4%2!=0 || y4%2!=0) continue; x3/=2; y3/=2; x4/=2; y4/=2; if (ok(x3,y3) && ok(x4,y4)) if (vis[x3][y3] && vis[x4][y4]) if (same(x3,y3,i,j) && same(x4,y4,i,j)) sum++; } printf("%d ",sum/2); } return 0; }