HDU6055-Regular polygon

Regular polygon

                                                                     Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
                                                                                               Total Submission(s): 685    Accepted Submission(s): 247

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：给你n个二维平面上的整数点，问这些整数点能形成几个正多边形

解题思路：因为是整数点，所以其实只能是正方形，然后可以枚举正方形的对角线，然后判断另外两点存不存在即可（一开始判断两点存不存在的数组开的不够大，出现越界，WA了）

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <algorithm>#include <map>#include <cmath>#include <set>#include <stack>#include <queue>#include <vector>#include <bitset>#include <functional>using namespace std;#define LL long longconst int INF = 0x3f3f3f3f;const LL mod = 1e9 + 7;struct point{    int x,y;} p[505];int n;int vis[205][205];int main(){    while(~scanf("%d",&n))    {        memset(vis,0,sizeof vis);        for(int i=1; i<=n; i++)        {            scanf("%d%d",&p[i].x,&p[i].y);            vis[p[i].x+100][p[i].y+100]=1;        }        int ans=0;        for(int i=1; i<=n; i++)        {            for(int j=i+1; j<=n; j++)            {                point tmp;                tmp.x=(p[i].x+p[j].x+p[j].y-p[i].y);                tmp.y=(p[i].y+p[j].y+p[i].x-p[j].x);                if(tmp.x%2!=0||tmp.y%2!=0) continue;                tmp.x/=2,tmp.y/=2;                if(tmp.x < -100 || tmp.x > 100 || tmp.y < -100 || tmp.y > 100 || !vis[tmp.x+100][tmp.y+100]) continue;                tmp.x=(p[i].x+p[j].x+p[i].y-p[j].y);                tmp.y=(p[i].y+p[j].y+p[j].x-p[i].x);                if(tmp.x%2!=0||tmp.y%2!=0) continue;                tmp.x/=2,tmp.y/=2;                if(tmp.x < -100 || tmp.x > 100 || tmp.y < -100 || tmp.y > 100 || !vis[tmp.x+100][tmp.y+100]) continue;                ans++;            }        }        printf("%d\n",ans/2);    }    return 0;}