H - H HDU - 5547（模拟）

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在三国时代背景下，Yi Sima设计了一种类似现代数独的游戏，但规模更小，要求玩家在4x4的网格中填入数字1到4，确保每一行、每一列以及四个2x2的子区域内的数字不重复。游戏中的某些数字被移除，玩家的任务是根据给定的线索恢复完整的数独板。

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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 4×44×4 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 2×22×2 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(1≤T≤100). TT 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 xx 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

题意：给你一个4*4的格子，*号代表未知数未知数只会是1 2 3 4，每一行，列的数字不能相同，把数列分成4部分，这四部分的数字也不能相同。

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
int mapp[10][10],flag=0;
int check (int x, int y, int s) // 判断是否有重复
{
    for(int i = 1; i <= 4; i++)//判断每行每列的数字是否重复
    {
        if(mapp[x][i] == s)
            return 0;
        if(mapp[i][y] == s)
            return 0;
    }
    if(x <= 2 && y <= 2)//判断四个部分中的数字是否重复
    {
        for(int i = 1; i <= 2; i++)
            for(int j = 1; j <= 2; j++)
                if(mapp[i][j] == s)
                    return 0;
    }
    if(x >= 3 && y <= 2)
    {
        for(int i = 3; i <= 4; i++)
            for(int j = 1; j <= 2; j++)
                if(mapp[i][j] == s)
                    return 0;
    }
    if(x >= 3 && y >= 3)
    {
        for(int i = 3; i <= 4; i++)
            for(int j = 3; j <= 4; j++)
                if(mapp[i][j] == s)
                    return 0;
    }
    if(x <= 2 && y>= 3)
    {
        for(int i = 1; i <= 2; i++)
            for(int j = 3; j <= 4; j++)
                if(mapp[i][j] == s)
                    return 0;
    }
    return 1;
}
void dfs(int x,int y)
{
    if(x>4)//如果大于4直接输出
    {
        for(int i=1; i<=4; i++)
        {
            for(int j=1; j<=4; j++)
            {
                printf("%d",mapp[i][j]);
            }
            printf("\n");
        }
        return;
    }
    if(mapp[x][y])//如果是数字直接跳过
    {
        if(y>=4)
            dfs(x+1,1);
        else
            dfs(x,y+1);
    }
    else
    {
        for(int i=1; i<=4; i++)//把四个数字都遍历一遍
        {
            if(check(x,y,i))//判断是否符合条件
            {
                mapp[x][y]=i;
                if(y>=4)
                    dfs(x+1,1);
                else
                    dfs(x,y+1);
                mapp[x][y]=0;//回溯
            }
        }
    }
}
int main()
{
    int t,k=1;
    scanf("%d",&t);
    while(t--)
    {
        char a;
        for(int i=1; i<=4; i++)
        {
            for(int j=1; j<=4; j++)
            {
                cin>>a;
                if(a!='*')//把字符转化为数字
                    mapp[i][j]=a-'0';
                else
                    mapp[i][j]=0;
            }
        }
        printf("Case #%d:\n",k++);
        flag=0;
        dfs(1,1);
    }
}