数独求解

// onlynum.cpp : 定义控制台应用程序的入口点。//#include "stdafx.h"#include <iostream>using namespace std;int _tmain(int argc, _TCHAR* argv[]){int a[9][9] = {{1, 3, 0, 9, 8, 0, 4, 0, 0},{0, 0, 5, 1, 3, 0, 0, 8, 9},{0, 0, 9, 0, 0, 7, 1, 0, 0},{3, 0, 2, 4, 5, 0, 6, 0, 7},{4, 5, 0, 0, 7, 0, 0, 0, 8},{7, 0, 0, 6, 0, 0, 3, 5, 0},{0, 2, 1, 7, 0, 3, 0, 0, 6},{8, 0, 3, 0, 0, 4, 9, 7, 0},{0, 0, 4, 0, 6, 1, 0, 2, 0}};int b[9][9][9];for (int i = 0; i < 9; i++){for (int j = 0; j < 9; j++){for (int k = 0; k < 9; k++)b[i][j][k] = k + 1;}}bool tag = true;while (tag){// 处理行for (int i = 0; i < 9; i++){for (int j = 0; j < 9; j++){if (a[i][j] != 0){for (int k = 0; k < 9; k++){b[i][k][a[i][j]-1] = 0;b[i][j][k] = 0;}}}}// 处理列for (int i = 0; i < 9; i++){for (int j = 0; j < 9; j++){if (a[i][j] != 0){for (int k = 0; k < 9; k++){b[k][j][a[i][j]-1] = 0;b[i][j][k] = 0;}}}}// 处理方块区for (int i = 0; i < 9; i++){for (int j = 0; j < 9; j++){if (a[i][j] != 0){for (int k = 0; k < 9; k++){b[i/3*3 + k/3][j/3*3 + k%3][a[i][j]-1] = 0;b[i][j][k] = 0;}}}}// 处理后，是否有新的确定的项tag = false;for (int i = 0; i < 9; i++){for (int j = 0; j < 9; j++){if (a[i][j] == 0){int count = 0, index = -1;;for (int k = 0; k < 9; k++){if (b[i][j][k] != 0){count++;index = k;}}if (count == 1){a[i][j] = index + 1;b[i][j][index] = 0;tag = true;// 还有优化的余地}}}}}// 打印结果for (int i = 0; i < 9; i++){for (int j = 0; j < 9; j++){cout << a[i][j] <<"/";for (int k = 0; k < 9; k++)if (b[i][j][k] != 0)cout << b[i][j][k];cout << "/t";}cout << endl;}return 0;}