题意：标准的一道数独题

DFS做法：

将横纵九宫格里的数字用位运算状态压缩，且可以通过逻辑或来确定总共有哪些数字被选择了，很方便也很快，代码如下

#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <ctime>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <string>
#define oo 0x13131313
using namespace std;
char MAP[30][30];
int ANS[30][30];
int x[30],y[30],z[30];
int nx[30]={0,1,1,1,4,4,4,7,7,7};
int ny[30]={0,1,4,7,1,4,7,1,4,7};
int nn[20][20];
void init()
{freopen("a.in","r",stdin);freopen("a.out","w",stdout);
}
void input()
{memset(x,0,sizeof(x));memset(y,0,sizeof(y));memset(z,0,sizeof(z));memset(ANS,0,sizeof(ANS));for(int i=2;i<=9;i++)gets(MAP[i]);gets(MAP[10]);for(int i=1;i<=9;i++)for(int j=0;j<=16;j=j+2)if(MAP[i][j]!='?') x[i]=(x[i]|(1<<MAP[i][j]-'0')),ANS[i][((j/2)+1)]=MAP[i][j]-'0',y[((j/2)+1)]=(y[((j/2)+1)]|(1<<(MAP[i][j]-'0')));for(int k=1;k<=9;k++)for(int i=nx[k];i<=nx[k]+2;i++)for(int j=ny[k];j<=ny[k]+2;j++){if(MAP[i][((j-1)*2)]!='?') z[k]=(z[k]|(1<<(MAP[i][((j-1)*2)]-'0')));nn[i][j]=k;}
}
int dfs(int X,int Y)
{int XX=X,YY=Y;if(YY+1<=9) YY++;else XX++,YY=1;if(X==10) return 1;else{if(ANS[X][Y]!=0) {if(dfs(XX,YY)) return 1;}else{int xxx=x[X],yyy=y[Y],zzz=z[nn[X][Y]];int t=x[X]|y[Y]|z[nn[X][Y]];for(int i=1;i<=9;i++){if((1&(t>>i))==0){ANS[X][Y]=i;x[X]=(x[X]|(1<<i));y[Y]=(y[Y]|(1<<i));z[nn[X][Y]]=(z[nn[X][Y]]|(1<<i));if(dfs(XX,YY)) return 1;ANS[X][Y]=0;x[X]=xxx;y[Y]=yyy;z[nn[X][Y]]=zzz;}}}}return 0;
}
void solve()
{dfs(1,1);for(int i=1;i<=9;i++){for(int j=1;j<=9;j++){printf("%d",ANS[i][j]);if(j!=9) printf(" ");}printf("\n");}
}
int main()
{// init();int Case=0;while(gets(MAP[1])){if(Case++) printf("\n");input();solve();}
}

DLX 做法待研究