17 多校

Regular polygon

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 122    Accepted Submission(s): 45

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

 

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=6055

题意：给你n个整点的坐标，问你能形成几个正多边形。

主要的就是点为整点，就是坐标的x和y均为整数。

官方题解：

容易得知只有正四边形可以使得所有的顶点为整数点。（具体证明可参考杨景钦在2017的国家队论文） 所以正解即求出所有的正四边形个数。

枚举2个点，然后暴力判断另外2个点的位置是否存在。

复杂度 N*N*logN。

#include<iostream>#include<string.h>#include<stdio.h>using namespace std;int n,mp[1022][1022],x[555],y[555];int main(){    while(~scanf("%d",&n))    {        memset(mp,0,sizeof mp);        for(int i=0;i<n;i++)        {            scanf("%d%d",&x[i],&y[i]);            x[i]+=350;            y[i]+=350;            mp[x[i]][y[i]] = 1;        }        int ans = 0;        int a ,b;        for(int i=0;i<n;i++)        {            for(int j=0;j<n;j++)            {                if(j==i)                    continue;                a = x[j] - x[i];                b = y[j] - y[i];                if(mp[x[i]-b][y[i]+a]&&mp[x[i]+a-b][y[i]+b+a])                    ans++;                if(mp[x[i]+b][y[i]-a]&&mp[x[i]+a+b][y[i]+b-a])                    ans++;            }        }        printf("%d\n",ans/8);    }    return 0;}