zoukankan      html  css  js  c++  java
  • POJ 2676 Sudoku

    Sudoku
    Time Limit: 2000MS   Memory Limit: 65536K
    Total Submissions: 12005   Accepted: 5984   Special Judge

    Description

    Sudoku is a very simple task. A square table with 9 rows and 9 columns is divided to 9 smaller squares 3x3 as shown on the Figure. In some of the cells are written decimal digits from 1 to 9. The other cells are empty. The goal is to fill the empty cells with decimal digits from 1 to 9, one digit per cell, in such way that in each row, in each column and in each marked 3x3 subsquare, all the digits from 1 to 9 to appear. Write a program to solve a given Sudoku-task. 

    Input

    The input data will start with the number of the test cases. For each test case, 9 lines follow, corresponding to the rows of the table. On each line a string of exactly 9 decimal digits is given, corresponding to the cells in this line. If a cell is empty it is represented by 0.

    Output

    For each test case your program should print the solution in the same format as the input data. The empty cells have to be filled according to the rules. If solutions is not unique, then the program may print any one of them.

    Sample Input

    1
    103000509
    002109400
    000704000
    300502006
    060000050
    700803004
    000401000
    009205800
    804000107

    Sample Output

    143628579
    572139468
    986754231
    391542786
    468917352
    725863914
    237481695
    619275843
    854396127
    题目大意:数独填空。
    解题方法:搜索。
    #include <stdio.h>
    #include <iostream>
    #include <string.h>
    using namespace std;
    
    typedef struct
    {
        int x;
        int y;
    }Point;
    
    Point p[85];
    
    char Maze[15][15];
    int nCount = 0;
    bool bfind = false;
    
    bool Judge1(int row, int n)
    {
        for (int i = 0; i < 9; i++)
        {
            if (Maze[row][i] == n)
            {
                return false;
            }
        }
        return true;
    }
    
    bool Judge2(int col, int n)
    {
        for (int i = 0; i < 9; i++)
        {
            if (Maze[i][col] == n)
            {
                return false;
            }
        }
        return true;
    }
    
    bool Judge3(int row, int col, int n)
    {
        row = row / 3;
        col = col / 3;
        for (int i = row * 3; i < row * 3 + 3; i++)
        {
            for (int j = col * 3; j < col * 3 + 3; j++)
            {
                if (Maze[i][j] == n)
                {
                    return false;
                }
            }
        }
        return true;
    }
    
    void DFS(int Step)
    {
        if (Step == nCount && !bfind)
        {
            bfind = true;
            for (int i = 0; i < 9; i++)
            {
                for (int j = 0; j < 9; j++)
                {
                    printf("%d", Maze[i][j]);
                }
                printf("
    ");
            }
        }
        for (int i = 1; i <= 9; i++)
        {
            if (Judge1(p[Step].x, i) && Judge2(p[Step].y, i) && Judge3(p[Step].x, p[Step].y, i) && !bfind && Maze[p[Step].x][p[Step].y] == 0)
            {
                Maze[p[Step].x][p[Step].y] = i;
                DFS(Step + 1);
                Maze[p[Step].x][p[Step].y] = 0;
            }
        }
    }
    
    int main()
    {
        int nCase;
        char str[10];
        scanf("%d", &nCase);
        memset(Maze, 0, sizeof(Maze));
        while(nCase--)
        {
            nCount = 0;
            bfind = false;
            for (int i = 0; i < 9; i++)
            {
                scanf("%s", str);
                for (int j = 0; j < 9; j++)
                {
                    Maze[i][j] = str[j] - '0';
                    if (Maze[i][j] == 0)
                    {
                        p[nCount].x = i;
                        p[nCount].y = j;
                        nCount++;
                    }
                }
            }
            DFS(0);
        }
        return 0;
    }
  • 相关阅读:
    NoSQL 数据库中的 CAP 理论
    JVM 相关概念
    Servlet 生命周期
    RabbitMQ
    消息队列概述
    05.类加载器深入解析及重要特性剖析
    LINUX笔记3(用户管理)
    LINUX笔记2(创建和编辑文本)
    修改httpd端口后,服务不能启动。
    LINUX笔记1(命令行和目录结构)
  • 原文地址:https://www.cnblogs.com/lzmfywz/p/3263506.html
Copyright © 2011-2022 走看看