poj2676Sudoku 暴搜

//求数独
//给出9*9的宫格，其中一些位置有1到9的数填住
//让填出没有剩下数使得每一行，每一列，每一个3*3的子格的数都是1—9
//直接爆搜，用r[i][j]，c[i][j],s[i][j]记录第行，列，子格是否有第j个数
#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std ;
const int maxn = 200 ;
char map[maxn][maxn] ;
int r[maxn][maxn];
int c[maxn][maxn];
int s[maxn][maxn];
int x[maxn] , y[maxn] ;
int len ;
int dfs(int pos)
{
 if(pos == len + 1)return 1;
 int u = x[pos] , v = y[pos] ,num = (u-1)/3*3+(v+2)/3;
 for(int i = 1;i <= 9;i++)
 {
 if(r[u][i]||c[v][i]||s[num][i])continue ;
 r[u][i] = c[v][i] = s[num][i] = 1;
 map[u][v] = i + '0';
 if(dfs(pos+1))return 1;
 map[u][v] = '0' ;
 r[u][i] = c[v][i] = s[num][i] = 0;
 }
 return 0 ;
}

int main()
{
 //freopen("in.txt","r",stdin) ;
 int T ;
 scanf("%d" , &T) ;
 while(T--)
 {
 len = 0 ;
 memset(r , 0 , sizeof(r)) ;
 memset(c , 0 , sizeof(c)) ;
 memset(s , 0 ,sizeof(s)) ;
 for(int i = 1;i <= 9;i++)
 {
 scanf("%s" ,&map[i][1]) ;
 for(int j = 1;j <= 9;j++)
 {
 if(map[i][j] == '0')x[++len]=i,y[len]=j;
 int t = map[i][j] - '0';
 r[i][t] = 1;
 c[j][t] = 1;
 int pos = (i-1)/3*3+(j+2)/3 ;
 s[pos][t] = 1;
 }
 }
 if(dfs(1) == 0){puts("-1");continue ;}
 for(int i = 1;i <= 9;i++)
 {
 for(int j = 1;j <= 9;j++)
 printf("%c",map[i][j]);
 puts("");
 }
 }
 return 0;
}