【城会玩系列】hdu 5365 Run【计算几何相关】

本文链接：https://blog.youkuaiyun.com/mengxiang000000/article/details/50654744

探讨一个关于在整数坐标系中寻找可以形成正三角形、正方形、正五边形或正六边形路径的问题。通过遍历所有可能的路径组合来计算满足条件的路径数量。

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Run

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1068 Accepted Submission(s): 471

Problem Description

AFA is a girl who like runing.Today,he download an app about runing .The app can record the trace of her runing.AFA will start runing in the park.There are many chairs in the park,and AFA will start his runing in a chair and end in this chair.Between two chairs,she running in a line.she want the the trace can be a regular triangle or a square or a regular pentagon or a regular hexagon.
Please tell her how many ways can her find.
Two ways are same if the set of chair that they contains are same.

Input

There are multiply case.
In each case,there is a integer n(1 < = n < = 20)in a line.
In next n lines,there are two integers xi,yi(0 < = xi,yi < 9) in each line.

Output

Output the number of ways.

Sample Input

Sample Output

很有意识的一个题、开始觉得是想把所有边的长度都求出来，然后判断凸包，判断线段走向、极角排序啥的，合计老多了，完事我觉得这个题正确率这么高，不应该这么复杂，带着好奇心和急切的找到正确思路，我找到了官方题解。。。然后就没有然后了。。。

他们是这样给的题解：地球人都知道，整数点是不能构成正3.5,6边形的。。。。

然后默默暴力、、

#include<stdio.h>
#include<algorithm>
#include<string.h>
using namespace std;
struct zuobiao
{
    int x,y;
}a[121];
int cal(int x,int y)
{
    return x*x+y*y;
}
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        for(int i=0;i<n;i++)
        {
            scanf("%d%d",&a[i].x,&a[i].y);
        }
        int output=0;
        for(int i=0;i<n;i++)
        {
            for(int j=i+1;j<n;j++)
            {
                for(int k=j+1;k<n;k++)
                {
                    for(int l=k+1;l<n;l++)
                    {
                        int d[6];
                        d[0]=cal(a[i].x-a[j].x,a[i].y-a[j].y);
                        d[1]=cal(a[i].x-a[k].x,a[i].y-a[k].y);
                        d[2]=cal(a[i].x-a[l].x,a[i].y-a[l].y);
                        d[3]=cal(a[j].x-a[k].x,a[j].y-a[k].y);
                        d[4]=cal(a[j].x-a[l].x,a[j].y-a[l].y);
                        d[5]=cal(a[k].x-a[l].x,a[k].y-a[l].y);
                        sort(d,d+6);
                        if(d[0]==d[1]&&d[1]==d[2]&&d[2]==d[3]&&d[4]==d[5]&&2*d[0]==d[4])
                        output++;
                    }
                }
            }
        }
        printf("%d\n",output);
    }
}