DFS（poj 2676）

题目：Sudoku

题意：求解数独。从样例和结果来看应该是简单难度的数独

思路：DFS

　　　设置3个数组，row[i][j] 判断第i行是否放了j数字，col[i][j] 判断第i列是否放了j数字。square[i/3][j/3][x]判断第i/3行第j/3列个宫是否放置了x数字；

#include <iostream>
#include <algorithm>
#include <stdlib.h>
#include <time.h>
#include <cmath>
#include <cstdio>
#include <string>
#include <cstring>
#include <vector>
#include <queue>
#include <stack>
#include <set>

#define c_false ios_base::sync_with_stdio(false); cin.tie(0)
#define INF 0x3f3f3f3f
#define INFL 0x3f3f3f3f3f3f3f3f
#define zero_(x,y) memset(x , y , sizeof(x))
#define zero(x) memset(x , 0 , sizeof(x))
#define MAX(x) memset(x , 0x3f ,sizeof(x))
#define swa(x,y) {LL s;s=x;x=y;y=s;}
using namespace std ;
#define N 50005
const double PI = acos(-1.0);
typedef long long LL ;

bool col[10][10], row[10][10], square[5][5][10];
char mapp[10];
int MAP[10][10];
int n;

bool dfs(int z){
    if(z>=81) return true;
    int x = z/9;
    int y = z%9;
    if(MAP[x][y])
        return dfs(z+1);
    for(int i = 1; i<= 9; i++){
        if(!row[x][i] && !col[y][i] && !square[x/3][y/3][i]){
            MAP[x][y] = i;
            row[x][i] = col[y][i] = square[x/3][y/3][i] = 1;
            if(dfs(z+1))
                return true;
            MAP[x][y] = 0;
            row[x][i] = col[y][i] = square[x/3][y/3][i] = 0;
        }
    }
    return false;
}
int main(){
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    scanf("%d", &n);
    while(n--){
        //memset(MAP, 0, sizeof(MAP));
        memset(row, false, sizeof(row));
        memset(col, false, sizeof(col));
        memset(square, false, sizeof(square));
        for(int i = 0; i<9; i++){
            scanf("%s", mapp);
            for(int j = 0; j<9; j++){
                MAP[i][j] = mapp[j] - ‘0‘;
                int c = MAP[i][j];
                if(MAP[i][j]) {
                    row[i][c] = col[j][c] = square[i/3][j/3][c] = 1;
                }
            }
        }
        dfs(0);
        for(int i=0;i<9;i++){
            for(int j=0;j<9;j++)
                printf("%d",MAP[i][j]);
            printf("\n");
        }
    }
    return 0;
}