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

1103000509002109400000704000300502006060000050700803004000401000009205800804000107

 

Sample Output

143628579572139468986754231391542786468917352725863914237481695619275843854396127

 

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！