Sudoku（简单DFS）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5547

数据比较少,直接暴力DFS,检验成立情况即可

AC代码：但是不知道为什么用scanf,printf输入输出就WA了

/* */
# include <iostream>
# include <stdio.h>
# include <string.h>
# include <cstdlib>
# include <cmath>
# include <climits>
# include <deque>
# include <queue>
# include <stack>
# include <vector>
# include <map>
# include <set>
# include <ctime>
# include <functional>
using namespace std;
typedef long long ll;
const int inf=0x3f3f3f3f;
const int maxn=5e4+50;
const ll mod=1e9+7;
char ma[25][25];
int vis[25];
int p[25][25];
int tx[20];
int ty[20];
int flag, tot;

int check()
{
    for(int i=0; i<4; i++ )///检查每一行
    {
        vis[1]=vis[2]=vis[3]=vis[4]=0;
        for(int j=0; j<4; j++ )
        {
            if( p[i][j]==0 )
                continue;
            if( vis[p[i][j]] )///填的数重复了不成立
                return 0;
            vis[p[i][j]] = 1;
        }
    }

    for(int j=0; j<4; j++ )///检查每一列
    {
        vis[1]=vis[2]=vis[3]=vis[4]=0;
        for(int i=0; i<4; i++ )
        {
            if( p[i][j]==0 )
                continue;
            if( vis[p[i][j]] )
                return 0;
                vis[p[i][j]] = 1;
        }
    }

    vis[1] = vis[2] = vis[3] = vis[4] = 0;
    for(int i=0; i<2; i++ )///检查左上角的1/4方块
    {
        for(int j=0; j<2; j++ )
        {
            if( p[i][j]==0 )
                continue;
            if( vis[p[i][j]] )
                return 0;
            vis[p[i][j]] = 1;
        }
    }

    vis[1]=vis[2]=vis[3]=vis[4]=0;
    for(int i=0; i<2; i++ )///检查右上角的1/4块
    {
        for(int j=2; j<4; j++ )
        {
            if( p[i][j]==0 )
                continue;
            if( vis[p[i][j]] )
                return 0;
            vis[p[i][j]] = 1;
        }
    }

    vis[1] = vis[2] = vis[3] = vis[4]=0;
    for(int i=2; i<4; i++ )///检查左下角1/4块
    {
        for(int j=0; j<2; j++ )
        {
            if( p[i][j]==0 )
                continue;
            if( vis[p[i][j]] )
                return 0;
            vis[p[i][j]] = 1;
        }
    }

    vis[1] = vis[2] = vis[3] = vis[4] = 0;
    for(int i=2; i<4; i++ )///检查右下角1/4块
    {
        for(int j=2; j<4; j++ )
        {
            if( p[i][j]==0 )
                continue;
            if( vis[p[i][j]] )
                return 0;
            vis[p[i][j]] = 1;
        }
    }
    return 1;
}


void dfs(int x)
{
    if( flag )
        return ;
    if( x==tot )
    {
        for(int i=0; i<4; i++ )
        {
            for(int j=0; j<4; j++ )
            {
                cout<<p[i][j];
                //printf("%d", p[i][j]);
            }
            cout<<endl;
            //printf("
");
        }
        flag=1;
        return ;
    }

    for(int i=1; i<=4; i++ )
    {
        p[tx[x]][ty[x]]=i;
        if( check())
        {
            dfs(x+1);
        }
        p[tx[x]][ty[x]] = 0;
    }
}

int main()
{
    int t;
    scanf("%d", &t);
    int k=1;
    //getchar();
    while( t-- )
    {
        for(int i=0; i<4; i++ )
        {
            for(int j=0; j<4; j++ )
            {
                cin>>ma[i][j];
                //scanf("%c", &ma[i][j]);
            }
            //getchar();
        }

        for(int i=0; i<4; i++ )
        {
            for(int j=0; j<4; j++ )
            {
                if( ma[i][j]=='*' )
                {
                    p[i][j] = 0;
                }
                else
                    p[i][j] = ma[i][j] - '0';
            }
        }

        flag=0;
        tot = 0;
        for(int i=0; i<4; i++ )
        {
            for(int j=0; j<4; j++ )
            {
                if( !p[i][j] )
                {
                    /*依次记录空节点的坐标*/
                    tx[tot] = i;
                    ty[tot] = j;
                    tot++;
                }
            }
        }
        printf("Case #%d:
", k++);
        dfs(0);
    }
    return 0;
}