HNU-离散数学-工具箱系列4-求等价类

帮助你输出离散数学作业上所有需要写等价类的过程与步骤

 求解如下这类问题：

代码如下：

#include<bits/stdc++.h>
using namespace std;
int main()
{
	cout<<"规模为n="<<endl; 
	int n;
	cin>>n;
	int m[n+1][n+1];
	cout<<"input matrix"<<endl;
	for (int i=1;i<=n;i++)
  	{
  	    for (int j=1;j<=n;j++)
  		{
  			cin>>m[i][j];
	 	 }
   	} //initialization
   	
   	
   	
   	//自反闭包r(R)
   	cout<<endl<<"以下为自反闭包r(R)"<<endl; 
   	int r[n+1][n+1];
   	for (int i=1;i<=n;i++)
  	{
  	    for (int j=1;j<=n;j++)
  		{
  			r[i][j]=m[i][j];
  			if (i==j) r[i][j]=1;
	 	 }
   	}//copy and prcess
   	for (int i=1;i<=n;i++)
  	{
  	    for (int j=1;j<=n;j++)
  		{
  			cout<<r[i][j]<<" ";
	 	 }
	 	 cout<<endl; 
   	}//output
   	cout<<endl<<endl;
   	 
   	 
   	 
	//对称闭包s(R) 
	cout<<endl<<"以下为对称闭包s(R) "<<endl; 
	int s[n+1][n+1];
   	for (int i=1;i<=n;i++)
  	{
  	    for (int j=1;j<=n;j++)
  		{
  			s[i][j]=r[i][j];
	 	 }
   	}//copy 
   	for (int i=1;i<=n;i++)
  	{
  	    for (int j=i;j<=n;j++)
  		{
  			if (s[i][j]==1 || s[j][i]==1)
  			{
  				s[i][j]=1;
  				s[j][i]=1;
			  }
	 	 }
   	}//process
   	for (int i=1;i<=n;i++)
  	{
  	    for (int j=1;j<=n;j++)
  		{
  			cout<<s[i][j]<<" ";
	 	 }
	 	 cout<<endl; 
   	}//output
   	cout<<endl<<endl;
   	
   	
   	
   	
   	
   	for (int i=1;i<=n;i++)
  	{
  	    for (int j=1;j<=n;j++)
  		{
  			m[i][j]=s[i][j];
	 	 }
	 	 cout<<endl; 
   	}//output
   	
   	
   	
   	
   	//可传递闭包t(R)
   	cout<<endl<<"以下为可传递闭包t(R) 及 其求解过程"<<endl; 
  	for (int j=1;j<=n;j++)
  	{
  		cout<<"("<<j<<")"<<"j="<<j<<",即第"<<j<<"列"<<endl; 
  		for (int i=1;i<=n;i++)
  		{
  			if (m[i][j]==1)
  			{
  				bool flag=false;
  				for (int k=1;k<=n;k++)
  				{
  					int temp=m[i][k];
  					m[i][k]=m[i][k]+m[j][k];
  					if (m[i][k]>1) m[i][k]=1;
  					if (temp!=m[i][k]) flag=true;
				  }
				cout<<"因第"<<i<<"行即M("<<i<<","<<j<<")=1,故第"<<i<<"行=第"<<i<<"行∨第"<<j<<"行，";
			    if (flag==true)
			    { 
			    	cout<<"得矩阵M"<<i<<","<<j<<"="<<endl;
					for (int i2=1;i2<=n;i2++)
  					{
  	  					  for (int j2=1;j2<=n;j2++)
  							{
  								cout<<m[i2][j2]<<" ";
	 						 }
	 						 cout<<endl;
   					}
   					cout<<endl;
				}
				else cout<<"保持不变"<<endl<<endl;
			  }
	 	 }
	 	 cout<<endl;
   	} 
	cout<<endl<<"以可传递闭包t(R) "<<endl;
	for (int i=1;i<=n;i++)
  	{
  	    for (int j=1;j<=n;j++)
  		{
  			cout<<m[i][j]<<" ";
	 	 }
	 	 cout<<endl; 
   	}
   	
	system("pause") ;
  return 0;
}