图的m着色问题 回溯法

#include <iostream>
#include <fstream>
using namespace std;

class Color
{
	friend int mColoring(int,int,int**);
	
	private:
		bool Ok(int k);
		void Backtrack(int t);

		int n,   //图的顶点数
		    m,   //可用颜色数 
			**a, //图的邻接矩阵
			*x;  //当前解
		long sum; //当前已找到的可m着色方案数 
}; 

bool Color::Ok(int k) //检查颜色的可用性 
{
	for(int j=1;j<=n;j++)
	{
		if((a[k][j]==1)&&(x[j]==x[k]))
		{
			return false;
		}
	} 
	
	return true;
} 

void Color::Backtrack(int t)
{
	if(t>n)
	{
		sum++;
		for(int i=1;i<=n;i++)
			cout<<x[i]<<" ";
		cout<<"\n";
	}else{
		for(int i=1;i<=m;i++)
		{
			x[t]=i;
			if(Ok(t))
				Backtrack(t+1);
			x[t]=0;
		}
	}
}

int mColoring(int n,int m,int **a)
{
	Color X;
	//初始化X
	X.n = n;
	X.m = m;
	X.a = a;
	X.sum = 0;
	int *p = new int [n+1];
	for(int i=0;i<=n;i++)
		p[i]=0;
	X.x = p;
	X.Backtrack(1); 
	delete [] p; 
	return X.sum; 
}

int main()
{
	ifstream input("input.txt");
	ofstream output("output.txt");
	
	int n,m,**arr;
	input>>n>>m;
	cout<<n<<" "<<m<<"\n";
	arr = new int* [n+1];
	for(int i=1;i<=n;i++)
	{
		*(arr+i) = new int [n+1];
	}
	
	for(int i=1;i<=n;i++)
	{
		for(int j=1;j<=n;j++)
		{
			input >> arr[i][j];
			cout << arr[i][j] << " ";
		}
		cout << "\n";
	}
	
	for(int i=1;i<=n;i++)
	{
		for(int j=1;j<=n;j++)
		{
			cout << arr[i][j] << " ";
			output << arr[i][j] << " ";
		}
		cout<<"\n";
		output<<"\n";
	}
	
	output << "\n"; 
	cout<<"\n";
	cout<<"\n"<<mColoring(n,m,arr)<<"种\n";
	
	return 0;
}