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  • 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(1T100)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
     
    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);
        }
    }
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  • 原文地址:https://www.cnblogs.com/upstart/p/8974882.html
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