Problem Description
A square is a 4-sided polygon whose sides have equal length and adjacent sides form 90-degree angles. It is also a polygon such that rotating about its centre by 90 degrees gives the same polygon. It is not the only polygon with the latter property, however, as a regular octagon also has this property.
So we all know what a square looks like, but can we find all possible squares that can be formed from a set of stars in a night sky? To make the problem easier, we will assume that the night sky is a 2-dimensional plane, and each star is specified by its x and y coordinates.
So we all know what a square looks like, but can we find all possible squares that can be formed from a set of stars in a night sky? To make the problem easier, we will assume that the night sky is a 2-dimensional plane, and each star is specified by its x and y coordinates.
Input
The input consists of a number of test cases. Each test case starts with the integer n (1 <= n <= 1000) indicating the number of points to follow. Each of the next n lines specify the x and y coordinates (two integers) of each point. You may assume that the points are distinct and the magnitudes of the coordinates are less than 20000. The input is terminated when n = 0.
Output
For each test case, print on a line the number of squares one can form from the given stars.
Sample Input
4
1 0
0 1
1 1
0 0
9
0 0
1 0
2 0
0 2
1 2
2 2
0 1
1 1
2 1
4
-2 5
3 7
0 0
5 2
0
Sample Output
1
6
1
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记着标记数组要用boolean类型
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1 #include<iostream> 2 #include<string> 3 #include<cstring> 4 #include<cmath> 5 #include<queue> 6 #include<cstdio> 7 using namespace std; 8 int x[1010],y[1010]; 9 bool fs[5000][5000];//此处要定义成bool类型的，否则超内存 10 int a,b,c,i,j,sum,n; 11 int f(int a,int b) 12 { 13 if(fs[a+2500][b+2500]) 14 return 1; 15 return 0; 16 } 17 18 int main() 19 { 20 int x1,y1,x2,y2,x3,y3,x4,y4; 21 int a,b,i,j,sum,n; 22 while(scanf("%d",&n)&&n) 23 { 24 sum=0; 25 memset(fs,0,sizeof(fs)); 26 for(i=0;i<n;i++) 27 { 28 scanf("%d %d",&a,&b); 29 x[i]=a; 30 y[i]=b; 31 fs[a+2500][b+2500]=1; 32 } 33 for(i=0;i<n;i++) 34 { 35 x1=x[i]; 36 y1=y[i]; 37 for(j=0;j<i;j++) 38 { 39 x2=x[j]; 40 y2=y[j]; 41 x3,y3,x4,y4; 42 x3=x1+(y1-y2);//这个公式推推就出来啦 43 y3=y1-(x1-x2); 44 x4=x2+(y1-y2); 45 y4=y2-(x1-x2); 46 if(f(x3,y3)&&f(x4,y4)) 47 sum++; 48 x3=x1-(y1-y2); 49 y3=y1+(x1-x2); 50 x4=x2-(y1-y2); 51 y4=y2+(x1-x2); 52 if(f(x3,y3)&&f(x4,y4)) 53 sum++; 54 } 55 } 56 printf("%d ",sum/4);//注意此处要除以4， 57 58 } 59 }