(搜索) poj 2676

时间：2015-02-12 19:57:37 阅读：172 评论：0 收藏：0 [点我收藏+]

标签：

M - Sudoku

Time Limit:2000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u

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.
技术分享

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

Sample Output

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<string>
#include<algorithm>
#include<cmath>
#include<queue>
using namespace std;
int tt,a[10][10],row[10][10],col[10][10],grid[10][10];
bool flag;
char s[10][10];
void print()
{
    for(int i=0;i<9;i++)
    {
        for(int j=0;j<9;j++)
        {
            printf("%d",a[i][j]);
        }
        printf("\n");
    }
}
void dfs(int x,int y)
{
    if(x==9)
    {
        flag=true;
        return ;
    }
    if(a[x][y]!=0)
    {
        if(y==8)
            dfs(x+1,0);
        else
            dfs(x,y+1);
    }
    else
    {
        int ret=3*(x/3)+y/3;
        for(int i=1;i<=9;i++)
        {
            if(row[x][i]||col[y][i]||grid[ret][i]) continue;
            a[x][y]=i;
            row[x][i]=1,col[y][i]=1,grid[ret][i]=1;
            if(y==8)
                dfs(x+1,0);
            else
                dfs(x,y+1);
            if(flag) return ;
            a[x][y]=0;
            row[x][i]=0,col[y][i]=0,grid[ret][i]=0;
        }
    }
}
int main()
{
    scanf("%d",&tt);
    while(tt--)
    {
        flag=false;
        memset(a,0,sizeof(a));
        memset(row,0,sizeof(row));
        memset(col,0,sizeof(col));
        memset(grid,0,sizeof(grid));
        for(int i=0;i<9;i++)
            scanf("%s",s[i]);
        for(int i=0;i<9;i++)
            for(int j=0;j<9;j++)
                a[i][j]=s[i][j]-‘0‘;
        for(int i=0;i<9;i++)
        {
            for(int j=0;j<9;j++)
            {
                if(a[i][j]!=0)
                {
                    row[i][a[i][j]]=1;
                    col[j][a[i][j]]=1;
                    grid[3*(i/3)+j/3][a[i][j]]=1;
                }
            }
        }
        dfs(0,0);
        print();
    }
    return 0;
}

(搜索) poj 2676

标签：

原文地址：http://www.cnblogs.com/a972290869/p/4288727.html

踩

(0)

评论一句话评论（0）

分享档案

更多>

2021年07月29日 (22)
2021年07月28日 (40)
2021年07月27日 (32)
2021年07月26日 (79)
2021年07月23日 (29)
2021年07月22日 (30)
2021年07月21日 (42)
2021年07月20日 (16)
2021年07月19日 (90)
2021年07月16日 (35)

周排行