HDU6055——多校2017

传送门：http://acm.hdu.edu.cn/showproblem.php?pid=6055

Regular polygon

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2777    Accepted Submission(s): 1110

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

 

Recommend

liuyiding

题意：在二维平面内有N个点，判断这n个点能组成多少个正多边形

思路：开始也很懵，不过后来知道坐标是整数点的正多边形只有正方形之后，思路清晰了很多。

贴上一份当时队友的代码：

#include<iostream>#include<sstream>#include<fstream>#include<set>#include<map>#include<cmath>#include<queue>#include<stack>#include<vector>#include<cstdio>#include<string>#include<cstring>#include<algorithm>using namespace std;#define si(x) scanf("%d", &x)#define sii(x,y) scanf("%d %d", &x, &y)#define sl(x) scanf("%ld", &x)#define sll(x,y) scanf("%ld %ld", &x, &y)#define fe1(i, a, b) for(int i = a; i <= b; i++)#define fu(i, a, b) for(int i = a; i < b; i++)#define MP(x, y) make_pair(x, y)typedef pair<int , int> pii;typedef long long LL;typedef unsigned long long LLu;const int maxn = 33;const int inf = 0x3f3f3f3f;using namespace std ;#define eqs 1e-9struct node{    double x , y ;}p[1100] ;bool cmp(node a,node b){    return ( a.x < b.x || ( a.x == b.x && a.y < b.y ) ) ;}bool judge(double x,double y,int n){    int low = 0 , mid , high = n-1 ;    while( low <= high )    {        mid = (low + high) / 2 ;        if( fabs(p[mid].x-x) < eqs && fabs(p[mid].y-y) < eqs )            return true ;        else if( p[mid].x-x > eqs || ( fabs(p[mid].x-x) < eqs && p[mid].y-y > eqs ) )            high = mid - 1 ;        else            low = mid + 1 ;    }    return false ;}int main(){    int n , i , j , num ;    double x , y , xx , yy ;    while(~scanf("%d", &n))    {        num = 0 ;        for(i = 0 ; i < n ; i++)        {            scanf("%lf %lf", &p[i].x, &p[i].y) ;        }        sort(p,p+n,cmp) ;        for(i = 0 ; i < n ; i++)        {            for(j = i+1 ; j < n ; j++)            {                if( i == j ) continue ;                x = (p[i].x+p[j].x)/2 ;                y = (p[i].y+p[j].y)/2 ;                xx = p[i].x - x ;                yy = p[i].y - y ;                if( judge(x+yy,y-xx,n) && judge(x-yy,y+xx,n) )                {                    num++ ;                }            }        }        printf("%d\n", num/2) ;    }    return 0;}