数独--暴力、剪枝搜索

import java.util.Scanner;
public class Main {
    static int arr[][];
    static Scanner in = new Scanner(System.in);
    public static void main(String[] args) {
        arr = new int[9][9];
        for (int i = 0; i < 9; i++) {
            for (int j = 0; j < 9; j++) {
                arr[i][j] = in.nextInt();
            }
        }
        dfs(0);
    }
    private static void dfs(int index) {
        if (index == 81) {
            for (int i = 0; i < 9; i++) {
                for (int j = 0; j < 9; j++) {
                    System.out.print(arr[i][j] + " ");
                }
                System.out.println();
            }
            return;
        }
        int row = index / 9;
        int col = index % 9;
        if (arr[row][col] == 0) {
            for (int i = 1; i <= 9; i++) {
                if (valid(i, row, col)) {
                    arr[row][col] = i;
                    dfs(index + 1);
                    arr[row][col] = 0;
                }
            }
        } else
            dfs(index + 1);
    }
    private static boolean valid(int num, int raw, int col) {
        for (int i = 0; i < 9; i++) {
            if ( arr[raw][i] == num||arr[i][col] == num) {
                return false;
            }
        }
        int tmpx = raw / 3;
        int tmpy = col / 3;
        for (int i = 0; i < 3; i++) {
            for (int j = 0; j < 3; j++) {
                if (arr[tmpx * 3 + i][tmpy * 3 + j] == num)
                    return false;
            }
        }
        return true;
    }
}

https://www.bilibili.com/video/av53554818

#include <iostream>
#include <algorithm>
using namespace std;
const int N = 9;
int ones[1 << N], map[1 << N];                //ones 表示 x 里面有几个1;//小抄 map表示 一个数最右边的1以及跟剩下的0组成后 为第几位
int row[N], col[N], cell[3][3];
char str[100];
inline int lowbit(int x)                    //lowbit 计算最低位！！！
{
    return x & -x;
}
void init()                                ////全部赋值位111111111 表示1~9个数都可以选择 简化后面的与运算
{
    for (int i = 0; i < N; i ++ ) row[i] = col[i] = (1 << N) - 1;
    for (int i = 0; i < 3; i ++ )
        for (int j = 0; j < 3; j ++ )
            cell[i][j] = (1 << N) - 1;
}
// 求可选方案的交集
inline int get(int x, int y)            ////进行与运算 表示真正可以选哪些数
{
    return row[x] & col[y] & cell[x / 3][y / 3];
}
bool dfs(int cnt)
{
    if (!cnt) return true;
    // 找出可选方案数最少的空格
    int minv = 10;
    int x, y;
    for (int i = 0; i < N; i ++ )
        for (int j = 0; j < N; j ++ )
            if (str[i * 9 + j] == ‘.‘)
            {
                int t = ones[get(i, j)];
                if (t < minv)
                {
                    minv = t;
                    x = i, y = j;
                }
            }
    for (int i = get(x, y); i; i -= lowbit(i))
    {
        int t = map[lowbit(i)];
        // 修改状态
        row[x] -= 1 << t;
        col[y] -= 1 << t;
        cell[x / 3][y / 3] -= 1 << t;
        str[x * 9 + y] = ‘1‘ + t;
        if (dfs(cnt - 1)) return true;
        // 恢复现场
        row[x] += 1 << t;
        col[y] += 1 << t;
        cell[x / 3][y / 3] += 1 << t;
        str[x * 9 + y] = ‘.‘;
    }
}
int main()
{
    for (int i = 0; i < N; i ++ ) map[1 << i] = i;
    for (int i = 0; i < 1 << N; i ++ )
    {
        int s = 0;
        for (int j = i; j; j -= lowbit(j)) s ++ ;
        ones[i] = s; // i的二进制表示中有s个1
    }
    while (cin >> str, str[0] != ‘e‘)
    {
        init();
        int cnt = 0;
        for (int i = 0, k = 0; i < N; i ++ )
            for (int j = 0; j < N; j ++ , k ++ )
                if (str[k] != ‘.‘)
                {
                    int t = str[k] - ‘1‘;
                    row[i] -= 1 << t;
                    col[j] -= 1 << t;
                    cell[i / 3][j / 3] -= 1 << t;
                }
                else cnt ++ ;
        dfs(cnt);
        cout << str << endl;
    }
    return 0;
}

原文地址：https://blog.51cto.com/14429166/2416831