poj2676(数独)

也是一个简单剪枝的dfs。记录所有为0的位置，依次填写，当发现某个空格可选的填写数字已经没有时，说明该支路无效，剪掉。

不算是一个难题吧，但是还是花了不少时间，问题主要出在细节上，行列坐标反了、3乘3小格的位置判断等。写程序一定要细心。

#include <iostream>
using namespace std;

const int MAX_R = ;
int map[MAX_R + ][MAX_R + ];
int zero_pos[MAX_R * MAX_R][];
int possible_digits[MAX_R * MAX_R][];
int zero_cnt;

int setPossibleDigits(int x, int y, int s)
{
    bool possible[];
    memset(possible, , sizeof(possible));
    for (int i = ; i <= MAX_R; i++){
        possible[map[i][x]] = true;
        possible[map[y][i]] = true;
    }
    int subX = (x - ) / ;
    int subY = (y - ) / ;
    for (int i = subX *  + ; i <= subX *  + ; i++){
        for (int j = subY *  + ; j <= subY *  + ; j++){
            possible[map[j][i]] = true;
        }
    }
    int cnt = ;
    for (int i = ; i <= ; i++){
        if (!possible[i])
            possible_digits[s][cnt++] = i;
    }
    return cnt;
}

bool dfs(int step)
{
    if (step == zero_cnt){
        return true;
    }
    int curX = zero_pos[step][],
        curY = zero_pos[step][];
    int possible_cnt = setPossibleDigits(curX, curY, step);
    if (possible_cnt == ){
        return false;
    }
    for (int i = ; i < possible_cnt; i++){
        map[curY][curX] = possible_digits[step][i];
        if (dfs(step + ))
            return true;
        else
            map[curY][curX] = ;
    }
    return false;
}

int main()
{
    int t;
    cin >> t;
    while (t--){
        memset(map, , sizeof(map));
        memset(zero_pos, , sizeof(zero_pos));
        memset(possible_digits, , sizeof(possible_digits));
        zero_cnt = ;
        for (int i = ; i <= MAX_R; i++){
            for (int j = ; j <= MAX_R; j++){
                char ch;
                cin >> ch;
                map[i][j] = ch - '';
                if (map[i][j] == ){
                    zero_pos[zero_cnt][] = i;
                    zero_pos[zero_cnt++][] = j;
                }
            }
        }
        dfs();
        for (int i = ; i <= MAX_R; i++){
            for (int j = ; j <= MAX_R; j++){
                cout << map[i][j];
            }
            cout << endl;
        }
    }
    return ;
}