poj Squares n个点，共能组成多少个正方形

 1 #include<iostream>
 2 #include<algorithm>
 3 #include<cstdio>
 4 #include<cstring>
 5 #include<cmath>
 6 using namespace std;
 7 int T;
 8 struct TT
 9 {
10     int x,y;
11 }node[1010];
12 bool cmp(TT a,TT b)
13 {
14     if(a.x<b.x ||(a.x==b.x && a.y<b.y))  return true;
15     return false;
16 }
17 bool judge(int x,int y)
18 {
19     int L = 1,R =T;
20    while(L<=R)
21     {
22        int mid = (L+R)/2;
23        if(node[mid].x == x && node[mid].y == y)
24        {
25            return true;
26        }
27        if(x == node[mid].x )   //这个地方老是出错，就是当x值相同的时候，会有两种情况的，我只是考虑了一种
28        {
29            if(y>node[mid].y) L = mid +1;
30                else  R = mid-1;
31        }
32        else if(x>node[mid].x) L = mid+1;
33          else                R = mid-1;
34     }
35     return false;
36 }
37 int main()
38 {
39    int ans;
40    int x1,y1,x2,y2,mx,my;
41    while(scanf("%d",&T) && T)
42    { ans = 0;
43        for(int i =1;i<=T;i++)
44           scanf("%d %d",&node[i].x,&node[i].y);
45           sort(node+1,node+T+1,cmp);//乱了一下，开始从0，开始排序了
46           for(int i  = 1; i<T; i++)
47             for(int j = i+1; j<=T; j++)
48             {
49               mx = node[j].x - node[i].x;
50               my = node[j].y - node[i].y;
51               x1 = node[i].x+my;
52               y1 = node[i].y-mx;
53               x2 = node[j].x+my;
54               y2 = node[j].y-mx;
55               if( judge(x1,y1) &&  judge(x2,y2))
56               {
57                     ans++;
58               }
59             }
60             printf("%d\n",ans/2);//因为每次都有两个重复；
61    }
62     return 0;
63 }