POJ 2002 二分 计算几何

根据正方形对角的两顶点求另外两个顶点公式：

x2 = (x1+x3-y3+y1)/2; y2 = (x3-x1+y1+y3)/2;

x4= (x1+x3+y3-y1)/2; y4 = (-x3+x1+y1+y3)/2;

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn=1000+5;
struct Node
{
    int x,y;
    bool operator<(const Node& rhs)const
    {
        return x<rhs.x || (x==rhs.x && y<rhs.y);
    }
}nodes[maxn];  

int main()
{
    int n;
    while(scanf("%d",&n)==1 && n)
    {
        int ans=0;//正方形个数
        for(int i=0;i<n;++i)
            scanf("%d%d",&nodes[i].x,&nodes[i].y);
        sort(nodes,nodes+n);  

        for(int i=0;i<n;++i)
        for(int j=i+1;j<n;++j)
        {
            Node tmp;//tmp作为正方形的第3或4个点
            tmp.x=nodes[i].x+nodes[i].y-nodes[j].y;
            tmp.y=nodes[i].y+nodes[j].x-nodes[i].x;
            if(!binary_search(nodes,nodes+n,tmp)) continue;
            tmp.x=nodes[j].x+nodes[i].y-nodes[j].y;
            tmp.y=nodes[j].y+nodes[j].x-nodes[i].x;
            if(!binary_search(nodes,nodes+n,tmp)) continue;
            ++ans;
        }
        printf("%d\n",ans/2);
    }
    return 0;
}