sudokuCrack

使用代码实例

至此我们有了一个稳健的求解方法，下面是一个实例

    #include<iostream>
    #include "sudokucrack.h"
    using namespace std;
    int main()
    {
        vvint d{//vvint见头文件
            {4,0,0,5,2,0,7,0,3},
            {0,0,0,0,0,3,0,0,0},
            {1,0,0,0,0,7,0,0,0},
            {0,1,4,0,0,0,0,0,6},
            {7,0,0,0,5,0,0,0,1},
            {5,0,0,0,0,0,4,2,0},
            {0,0,0,4,0,0,0,0,5},
            {0,0,0,8,0,0,0,0,0},
            {2,0,1,0,7,6,0,0,8}
        };
        SudokuCrack s(d);
        int m=s.beginCrack();
        if(m==0)
       	{
        	const vvint &r=s.getResult();
            for(const vint & a : r)//c++11，基于范围的for循环
            {
            	for( int d : a)
                	cout<<d<<" ";
                cout<<"\n";
            }
            cout<<endl;
        }
        else if(m==1)
        	cout<<"数独非法！在某行中有重复数字！"<<endl;
        else if(m==2)
        	cout<<"数独非法！在某列中有重复数字！"<<endl;
        else if(m==3)
        	cout<<"数独非法！在某3×3矩阵中有重复数字！"<<endl;
        return 0;
    }

瞬间完成，输出如下

    4 9 6 5 2 8 7 1 3 
    8 7 5 1 4 3 6 9 2 
    1 3 2 9 6 7 8 5 4 
    3 1 4 2 8 9 5 7 6 
    7 2 9 6 5 4 3 8 1 
    5 6 8 7 3 1 4 2 9 
    6 8 7 4 9 2 1 3 5 
    9 4 3 8 1 5 2 6 7 
    2 5 1 3 7 6 9 4 8