回溯算法（解任意阶数独）

<p>回溯算法的基本框架为</p><p>     函数名（int cnt）{   </p><p>     for()</p><p>        {</p><p>           赋值;</p><p>       if（==）｛</p><p>              ｝else{</p><p>                   函数名（cnt+1);</p><p>              }</p><p>           抹去; </p><p>        }</p><p>}</p>/*
 theme:求解数独
       回溯算法
  Coder:瞿鹏志
  time:2015.1.11
 */
#include <iostream>
using namespace std;
#define N 9
#include <math.h>
class suduk{
	private:
		int sudu[N][N];
		public:
			suduk();
			void SetSudk();//输入数独矩阵
		   bool Isvaild(int i,int j);
			void answer(int cnt);
			void Cout();
};
int main(void){
	suduk qus1;
	qus1.SetSudk();
    qus1.Cout();
    qus1.answer(0);
	return 0;
}
void suduk::Cout()
{
	for(int prin=0;prin<N*N;prin++){
	if(prin%N == 0) {
	 cout<<endl;
	}
	cout<<sudu[prin/N][prin%N]<<" ";
	}
	cout<<endl;
}
suduk::suduk(){
	for(int i=0;i<N*N;i++){
		sudu[i/N][i%N]=0;
	}
}
void suduk::SetSudk()
{
	int sudo[N][N]={{8,0,0,1,3,7,0,0,0},{6,0,0,9,0,0,0,1,0},{5,0,0,0,0,0,0,3,0},
	{0,0,0,3,8,0,0,0,9},{0,5,0,0,0,0,0,0,0},{9,0,0,0,0,0,8,7,0},{0,2,0,0,0,0,0,0,0},
	{0,0,0,0,0,6,2,4,3},{1,0,0,0,5,0,9,0,0}};
    for(int i=0;i<N;i++){
    	for(int j=0;j<N;j++){
    		sudu[i][j]=sudo[i][j];
    	//cin>>sudu[i][j];
		}
    }
}
bool suduk::Isvaild(int i,int j)
{
	int run;
    for(run=0;run<N;run++){
    	if((run!=j)&&sudu[i][run]==sudu[i][j]){
    		return false;
    	}
    	if((run!=i)&&sudu[run][j]==sudu[i][j]){
    		return false;
    	}
    }
    int jie=(int)pow((double)N,1.0/2.0);
    int row=i/jie*jie,col=j/jie*jie;
     for(run=0;run<N;run++){
     	if(row+run/jie!=i || col+run%jie != j){
     		if(sudu[row+run/jie][col+run%jie]==sudu[i][j]){
     			return false;
     		}
     	}
     }
	return true;
}
void suduk::answer(int cnt)
{
	int i=cnt/N;
	int j=cnt%N;
	if(sudu[i][j]==0){
		for(int num=1;num<=N;num++){
			sudu[i][j]=num;
			if(Isvaild(i,j)){
				if(cnt!=N*N-1){
					answer(cnt+1);
				}else{
					Cout();
				}
			}
           			sudu[i][j]=0;
		}
	}else{
		answer(cnt+1);
	}
}

回溯算法的基本框架为

函数名（int cnt）{

for()

{

if（）｛

｝else{

函数名（cnt+1);

}

}

}