POJ 2002 Squares

解题思路:

枚举两个顶点作为对角边，计算正方形另一对角边的两个顶点，查找这两个顶点是否都存在

此思路中每个正方形均会计算两次，所以最后结构需要/2

#include <iostream>
using namespace std;
#define PRIME 39997
#define MAXN 1000
int point[MAXN][2],first[PRIME],next[MAXN];
inline int CalHash(int a[2])
{
    int ans = a[0]*a[0]+a[1]*a[1];
    ans%=PRIME;
    return ans;
}
inline bool Find(int a[2])
{
    int hash,i;
    hash = CalHash(a);
    for(i=first[hash];i!=-1;i=next[i])
        if(point[i][0]==a[0]&&point[i][1]==a[1])return true;
    return false;
}
int main()
{
    int i,j,n,hash,t,ans,a[2],b[2];
    while(scanf("%d", &n) && n)
    {
        for(i=0;i<PRIME;i++)first[i]=-1;
        for(i=0;i<MAXN;i++)next[i]=-1;
        for (i=0;i<n;i++)
        {
            scanf("%d %d", &point[i][0], &point[i][1]);
            hash = CalHash(point[i]);
            next[i]=first[hash],first[hash]=i;
        }
        for (ans=0,i=1;i<n;i++)
        {
            for(j=0;j<i;j++)
            {
                t=point[i][0]+point[j][0]-point[i][1]+point[j][1];
                if(t%2)continue; a[0]=t/2;
                t=point[i][0]+point[j][0]+point[i][1]-point[j][1];b[0]=t/2;
                t=point[i][0]-point[j][0]+point[i][1]+point[j][1];a[1]=t/2;
                t=point[i][1]+point[j][1]-point[i][0]+point[j][0];b[1]=t/2;
                if(Find(a)&&Find(b))ans++;
            }
        }
        printf("%d\n",ans/2);
    }
    return 0;
}