hdu6055 Regular polygon【思维+几何基础】

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=6055
题意：给你n个坐标为整数的点，让你求有有多少个正多边形
解析：首先这个正多边形一定是正四边形，因为点是整点，接下来需要判断有几个，那么枚举两个点，然后去找这条线段构成的正方形是否存在，找正方形，用坐标旋转即可（90度很特殊，算起来很快）

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <algorithm>#include <bitset>#include <cmath>#include <vector>#include <queue>#include <set>using namespace std;typedef long long LL;struct point{    int x,y;}a[505];int vis[1000][1000];bool shun(point p0,point p1){    int x1 = p1.y-p0.y+p1.x;    int y1 = p0.x-p1.x+p1.y;    int x2 = p1.y-p0.y+p0.x;    int y2 = p0.x-p1.x+p0.y;    return vis[x1+500][y1+500]&&vis[x2+500][y2+500];}bool ni(point p0,point p1){    int x1 = p0.y-p1.y+p0.x;    int y1 = p1.x-p0.x+p0.y;    int x2 = p0.y-p1.y+p1.x;    int y2 = p1.x-p0.x+p1.y;    return vis[x1+500][y1+500]&&vis[x2+500][y2+500];}int main(){    int n;    while(~scanf("%d",&n))    {        memset(vis,0,sizeof(vis));        for(int i=0;i<n;i++)        {            scanf("%d %d",&a[i].x,&a[i].y);            vis[a[i].x+500][a[i].y+500] = 1;        }        int ans = 0;        for(int i=0;i<n;i++)        {            for(int j=i+1;j<n;j++)            {                if(shun(a[i],a[j]))                    ans++;                if(ni(a[i],a[j]))                    ans++;            }        }        printf("%d\n",ans/4);    }    return 0;}/*90 00 10 21 01 11 22 02 12 2*/