问题 B: 数独（九宫格经典问题+dfs）

5????7??6
?6????5?4
?834?????
???182?4?
??1???9??
?7?369???
?????543?
1?5????9?
7??2????1

514927386
967831524
283456179
659182743
321574968
478369215
892615437
135748692
746293851

思路：dfs。符合条件继续搜，不符合递推回来。

#pragma GCC optimize(3 , "Ofast" , "inline")#include <bits/stdc++.h>#define rep(i , a , b) for(register int i=(a);i<=(b);i++)
#define per(i , a , b) for(register int i=(a);i>=(b);i--)using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int , int> pi;template<class T>
inline void read (T &x) {x = 0;int sign = 1;char c = getchar ();while (c < '0' || c > '9') {if ( c == '-' ) sign = - 1;c = getchar ();}while (c >= '0' && c <= '9') {x = x * 10 + c - '0';c = getchar ();}x = x * sign;
}const int maxn = 2e5 + 10;
const int inf = int (1e9);
const ll INF = ll (1e18);
const double PI = acos (- 1);
const int mod = 1e9+7;
const double eps = 1e-8;char s[20][20];
bool visc[10][10];
bool visr[10][10];
bool visg[10][10];
int ans[20][20];bool dfs(int x,int y) {if(x==10) return true;bool f=0;if(ans[x][y]) {if(y==9) f=dfs (x+1,1);else f=dfs (x,y+1);return f;}int cnt = (x-1)/3*3+(y-1)/3+1;rep (i,1,9) {if(!visc[x][i]&&!visr[y][i]&&!visg[cnt][i]) {visc[x][i]=1;visr[y][i]=1;visg[cnt][i]=1;ans[x][y]=i;if(y==9) f=dfs (x+1,1);else f=dfs (x,y+1);if(f) return f;visc[x][i]=0;visr[y][i]=0;visg[cnt][i]=0;ans[x][y]=0;}}return false;
}
int main () {rep (i,1,9) {scanf ("%s",s[i]+1);rep (j,1,9) {if(s[i][j]=='?') {ans[i][j]=0;continue;}else ans[i][j]=s[i][j]-'0';visc[i][ans[i][j]]=1;visr[j][ans[i][j]]=1;visg[(i-1)/3*3+(j-1)/3+1][ans[i][j]]=1;}}dfs(1,1);rep (i,1,9) {rep (j,1,9) {printf ("%d",ans[i][j]);}printf ("\n");}return 0;
}