Sudoku (深搜)

Sudoku

Time Limit : 4000/2000ms (Java/Other)   Memory Limit : 131072/65536K (Java/Other)
Total Submission(s) : 30   Accepted Submission(s) : 14
Special Judge
Problem 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
 

Source
PKU
 

题意:

补充完整表格,使每一行每一列都含有1~9,并且在每个小方块内都有1~9,。

思路:

搜索题,关键是对小方格内的判断,此处先寻找小方格的最左上角然后开始检查!

代码:

#include <iostream>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
using namespace std;
int num,flag;
int mp[10][10];
char str[10];
struct point
{
    int x;
    int y;
}node[100];
void print()   //单独设立一个输出函数进行输出
{
    int i,j;
    for(i=0;i<9;i++)
    {
        for(j=0;j<8;j++)
        cout<<mp[i][j];
        cout<<mp[i][8]<<endl;
    }
}
bool judge(int xx,int yy)  //判断地几个不存在数的点它的行列和小方格内是否都满足
{
    int i,j;
    int a,b;
    for(i=0;i<9;i++)
    {
        if(mp[node[yy].x][i]==xx||mp[i][node[yy].y]==xx)
            return 0;
    }
    a=(node[yy].x/3)*3; //寻求小方格起点
    b=(node[yy].y/3)*3;
    for(i=a;i<a+3;i++)
        for(j=b;j<b+3;j++)
    {
          if(mp[i][j]==xx)
        return 0;
    }
    return 1;

}
void dfs(int k)
{
    if(k==num)
    {
        flag=1;
        print();
        return;
    }
    else
    {
        for(int i=1;i<=9;i++)
        {
            if(judge(i,k)&&!flag)
            {
                mp[node[k].x][node[k].y]=i;
                dfs(k+1);
                mp[node[k].x][node[k].y]=0;
            }
        }
    }
}
int main()
{
    int t;
    int i,j;
    cin>>t;
    while(t--)
    {
        num=0;
        for(i=0;i<9;i++)
         {
            cin>>str;
            for(j=0;j<9;j++)
            {
                if(str[j]=='0')
                {
                    node[num].x=i;
                    node[num++].y=j;
                    mp[i][j]=0;
                }
                else
                    mp[i][j]=str[j]-'0';

            }
         }
         flag=0;
         dfs(0);
    }
}

心得:

细心!

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