怎么判断由4个顶点连接成的图形是正方形?图中有多少个正方形?

//dev-a_square ，关于判断是否为正方形的函数。
#include <cstdlib>
#include <iostream> using namespace std; struct ve// square.vertex
{
int x;
int y;
};
bool a_square(ve a,ve b,ve c,ve d)
{
ve e;
//寻找a点的对角点，只要bcd中有一点与a不在同一直线上即可
if((a.x!=b.x )&&(a.y!=b.y))
{
cout<<"b may be a diagonal"<<endl;
//为了方便后续比较另外两个顶点与与这两个顶点的关系，
//将对角与b交换位置，此处正好是b所以无需交换
}
else
{
if((a.x!=c.x )&&(a.y!=c.y))
{
cout<<"c may be a diagonal"<<endl;
e=c;//c与b的位置交换
c=b;
b=e;
}
else
{
if((a.x!=d.x )&&(a.y!=d.y))
{
cout<<"d may be a diagonal"<<endl;
e=d;
d=b;
b=e;
}
else
{
// cout<<"abcd不可能组成正方形";
return 0;
}
}
}
//下面来判断cd和ab能否组成正方形。
if( ((c.x==a.x) && (c.y==b.y) //c和d的其中一个点的x为a的x，y为b的y
&&(d.x==b.x) && (d.y==a.y)) //另一个点的x为b的x，y为a的y
|| ((d.x==a.x) && (d.y==b.y)
&&(c.x==b.x) && (c.y==a.y)) )
{
//cout<<"abcd为正方形"<<endl;
return 1;
}
else
{
// cout<<"abcd不可能组成正方形";
return 0;
}
}
int main(int argc, char *argv[])
{
ve array[4]={{2,4},{4,4},{4,2},{2,2}};//测试
a_square(array[0],array[1],array[2],array[3]);
system("PAUSE");
return EXIT_SUCCESS;
}

//dev_c++，关于让所有可能的排列方式的实现
#include <cstdlib>
#include <iostream> using namespace std; int main(int argc, char *argv[])
{
int n=0,t=41;//n记录所有的组合，t为元素个数
for(int a=0;a<t;a++)
{
for (int b=0;b<t;b++)
{
for(int c=0;c<t;c++)
{
for(int d=0;d<t;d++)
{
//cout<< a<<"_"<<b<<"_"<<c<<"_"<<d<<endl;
n++;
}
}
}
}
cout<<t<<"中取4有"<<n<<"种排列"<<endl;
system("PAUSE");
return EXIT_SUCCESS;
}

再将这两个片断整合，求出可以组合成正方形的数N,然后用N/24就可以排除所有的重复计数的了，大功告成。后续我会贴出整合后的代码。请各位大师不吝指教。

//dev-a_square
#include <cstdlib>
#include <iostream>
#include <math.h> using namespace std; struct ve// square.vertex
{
int x;
int y;
};
ve array[41]={ //以左下角为原点,构造矩阵
{0,0},{2,0},{4,0},{6,0},{8,0},
{0,2},{2,2},{4,2},{6,2},{8,2},
{0,4},{2,4},{4,4},{6,4},{8,4},
{0,6},{2,6},{4,6},{6,6},{8,6},
{0,8},{2,8},{4,8},{6,8},{8,8},
{3,1},{3,2},{3,3},{3,5},{3,6},{3,7},
{4,1},{4,3},{4,5},{4,7},
{5,1},{5,2},{5,3},{5,5},{5,6},{5,7}
};
int n=0,n_end=0; bool a_square(ve a,ve b,ve c,ve d)
{
ve e;
//寻找a点的对角点，只要bcd中有一点与a不在同一直线上即可
if((a.x!=b.x )&&(a.y!=b.y))
{
// cout<<"b may be a diagonal"<<endl;
//为了方便后续比较另外两个顶点与与这两个顶点的关系，
//将对角与b交换位置，此处正好是b所以无需交换
}
else
{
if((a.x!=c.x )&&(a.y!=c.y))
{
//cout<<"c may be a diagonal"<<endl;
e=c;//c与b的位置交换
c=b;
b=e;
}
else
{
if((a.x!=d.x )&&(a.y!=d.y))
{
// cout<<"d may be a diagonal"<<endl;
e=d;
d=b;
b=e;
}
else
{
// cout<<"abcd不可能组成正方形";
return 0;
}
}
}
//下面来判断cd和ab能否组成正方形。
if( ((c.x==a.x) && (c.y==b.y) //c和d的其中一个点的x为a的x，y为b的y
&&(d.x==b.x) && (d.y==a.y)) //另一个点的x为b的x，y为a的y
|| ((d.x==a.x) && (d.y==b.y)
&&(c.x==b.x) && (c.y==a.y)) )
{
//cout<<"abcd为矩形"<<endl;
}
if( //这一句是最关键的,貌似就错在这里了,哪位大侠帮忙看看...
(abs(a.x-b.x))==(abs(d.y-c.y))//相邻的两条边长度相等 的矩形是正方形
||(abs(a.y-b.y))==(abs(c.x-d.x))
)
{
/*******
cout<<"("<<a.x<<","<<a.y<<")"<<" and "<<
"("<< b.x<<","<<b.y<<")"<<" and "<<
"("<<c.x<<","<<c.y<<")"<<" and "<<
"("<< d.x<<","<<d.y<<")"<<" 四点是图中的正方形之一 "<< endl;
*********/
return 1;
}
else
{
// cout<<"abcd不可能组成正方形";
return 0;
}
}
int main(int argc, char *argv[])
{
int t=41;//t=sizeof(array)/sizeof(ve)
for(int a=0;a<t;a++)
{
for (int b=0;b<t;b++)
{
for(int c=0;c<t;c++)
{
for(int d=0;d<t;d++)
{
if(a_square(array[a],array,array[c],array[d]))
{
n++;
}
}
}
}
}
n_end=(n/24)-1;//扣掉 (3,3),(3,5),(5,3),(5,5)这个没有连通的.
cout<<"这个图中有"<<n_end<<"个正方形"<<endl;
system("PAUSE");
return EXIT_SUCCESS;
}

//dev-a_square
#include <cstdlib>
#include <iostream>
#include <math.h>using namespace std;struct ve// square.vertex
{
int x;
int y;
};
ve array[41]={ //以左下角为原点,构造矩阵
{0,0},{2,0},{4,0},{6,0},{8,0},
{0,2},{2,2},{4,2},{6,2},{8,2},
{0,4},{2,4},{4,4},{6,4},{8,4},
{0,6},{2,6},{4,6},{6,6},{8,6},
{0,8},{2,8},{4,8},{6,8},{8,8},
{3,1},{3,2},{3,3},{3,5},{3,6},{3,7},
{4,1},{4,3},{4,5},{4,7},
{5,1},{5,2},{5,3},{5,5},{5,6},{5,7}
};
int n=0,n_end=0; bool a_square(ve a,ve b,ve c,ve d)
{
ve e;
if(//四个点中任意两个点都不能相同,否者就不够4个顶点了.
( (a.x==b.x) &&(a.y==b.y) ) ||
( (a.x==c.x) &&(a.y==c.y) ) ||
( (a.x==d.x) &&(a.y==d.y) ) ||
( (c.x==b.x) &&(c.y==b.y) ) ||
( (d.x==b.x) &&(d.y==b.y) ) ||
( (c.x==d.x) &&(c.y==d.y) )
)
{
return 0;
}
//寻找a点的对角点，只要bcd中有一点与a不在同一直线上即可
if((a.x!=b.x )&&(a.y!=b.y))
{
// cout<<"b may be a diagonal"<<endl;
//为了方便后续比较另外两个顶点与与这两个顶点的关系，
//将对角与b交换位置，此处正好是b所以无需交换
}
else
{
if((a.x!=c.x )&&(a.y!=c.y))
{
//cout<<"c may be a diagonal"<<endl;
e=c;//c与b的位置交换
c=b;
b=e;
}
else
{
if((a.x!=d.x )&&(a.y!=d.y))
{
// cout<<"d may be a diagonal"<<endl;
e=d;
d=b;
b=e;
}
else
{
// cout<<"abcd不可能组成正方形";
return 0;
}
}
}
//下面来判断cd和ab能否组成矩形。
if( ((c.x==a.x) && (c.y==b.y) //c和d的其中一个点的x为a的x，y为b的y
&&(d.x==b.x) && (d.y==a.y)) //另一个点的x为b的x，y为a的y
|| ((d.x==a.x) && (d.y==b.y)
&&(c.x==b.x) && (c.y==a.y)) )
{
//cout<<"abcd为矩形"<<endl;
}
else
{
return 0;
}
if( //验证临边是否相等
(abs(a.x-b.x)) ==(abs(a.y-b.y))//a点和b点是对角
)
{
/*******
cout<<"("<<a.x<<","<<a.y<<")"<<" and "<<
"("<< b.x<<","<<b.y<<")"<<" and "<<
"("<<c.x<<","<<c.y<<")"<<" and "<<
"("<< d.x<<","<<d.y<<")"<<" 四点是图中的正方形之一 "<< endl;
/*********/
return 1;
}
else
{
// cout<<"abcd不可能组成正方形";
return 0;
}
}
int main(int argc, char *argv[])
{
int t=41;//t=sizeof(array)/sizeof(ve)
for(int a=0;a<t;a++)
{
for (int b=0;b<t;b++)
{
for(int c=0;c<t;c++)
{
for(int d=0;d<t;d++)
{
if(a_square(array[a],array,array[c],array[d]))
{
n++;
}
}
}
}
}
n_end=(n/24)-1;//扣掉 (3,3),(3,5),(5,3),(5,5)这个没有连通的.
cout<<"这个图中有"<<n_end<<"个正方形"<<endl;
system("PAUSE");
return EXIT_SUCCESS;
}