The 2015 China Collegiate Programming Contest Sudoku

Sudoku

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 3015    Accepted Submission(s): 1010

Problem Description

Yi Sima was one of the best counselors of Cao Cao. He likes to play a funny game himself. It looks like the modern Sudoku, but smaller.

Actually, Yi Sima was playing it different. First of all, he tried to generate a 4×4 board with every row contains 1 to 4, every column contains 1 to 4. Also he made sure that if we cut the board into four 2×2 pieces, every piece contains 1 to 4.

Then, he removed several numbers from the board and gave it to another guy to recover it. As other counselors are not as smart as Yi Sima, Yi Sima always made sure that the board only has one way to recover.

Actually, you are seeing this because you've passed through to the Three-Kingdom Age. You can recover the board to make Yi Sima happy and be promoted. Go and do it!!!

 

Input

The first line of the input gives the number of test cases, T(1≤T≤100). T test cases follow. Each test case starts with an empty line followed by 4 lines. Each line consist of 4 characters. Each character represents the number in the corresponding cell (one of '1', '2', '3', '4'). '*' represents that number was removed by Yi Sima.

It's guaranteed that there will be exactly one way to recover the board.

 

Output

For each test case, output one line containing Case #x:, where x is the test case number (starting from 1). Then output 4 lines with 4 characters each. indicate the recovered board.

 

Sample Input

3 ****

2341

4123

3214

*243

*312

*421

*134

*41*

**3*

2*41

4*2*

 

Sample Output

Case #1:

1432

2341

4123

3214

Case #2:

1243

4312

3421

2134

Case #3:

3412

1234

2341

4123

 

Source

The 2015 China Collegiate Programming Contest

 

Recommend

wange2014

 

 

这个题唯一的问题，可能是需要思考一下i/2，y/2来划分四个格子

#include<bits/stdc++.h>
using namespace std;
#define rap(i,n) for(int i=0;i<n;i++)
#define dap(i,n) for(int i=1;i<=n;i++)
char a[10][10];

int check(int x,int b,char c)
{
    int i;
    rap(i,4)
    {
        if(a[x][i]==c||a[i][b]==c)
        return 1;
    }
    if(x/2==0&&b/2==0)
    {
        if(a[0][0]==c||a[0][1]==c||a[1][0]==c||a[1][1]==c)
        return 1;
    }
    else if(x/2==0&&b/2)
    {
        if(a[0][2]==c||a[0][3]==c||a[1][2]==c||a[1][3]==c)
        return 1;
    }
    else if(x/2&&b/2==0)
    {
        if(a[2][0]==c||a[2][1]==c||a[3][0]==c||a[3][1]==c)
        return 1;
    }
    else if(x/2&&b/2)
    {
        if(a[2][2]==c||a[2][3]==c||a[3][2]==c||a[3][3]==c)
        return 1;
    }
    return 0;
}
void dfs(int x,int y)
{
    int i,j;
    if(x==4)
    {
        for(i=0;i<4;i++)
        {
            for(j=0;j<4;j++)
            printf("%c",a[i][j]);
            printf("
");
        }
        return;
    }
    if(a[x][y]!='*')
    {
        if(y==3)
        dfs(x+1,0);
        else
        dfs(x,y+1);
    }
    else
    {
        for(i=1;i<=4;i++)
        {
            if(check(x,y,i+'0')==0)
            {
                a[x][y]=i+'0';
                if(y==3)
                dfs(x+1,0);
                else
                dfs(x,y+1);
                a[x][y]='*';
            }
        }

    }
}
int main()
{
    int t,cnt=0;
    cin>>t;
    while(t--)
    {
        rap(i,4)
        scanf("%s",&a[i]);
        cout<<"Case #"<<++cnt<<":"<<endl;
        dfs(0,0);
    }
}