poj2676 Sudoku

题意：

    数独游戏。

思路：

    深搜。

实现：

#include <iostream>
#include <cstdio>
using namespace std;
int n;
int now[9][9], row[9][10], col[9][10], block[3][3][10];
bool flag = false;

bool check(int x, int y, int i)
{
    if (!row[x][i] && !col[y][i] && !block[x / 3][y / 3][i])
    {
        row[x][i] = col[y][i] = block[x / 3][y / 3][i] = 1;
        return true;
    }
    return false;
}

void back_trace(int x, int y, int i)
{
    now[x][y] = row[x][i] = col[y][i] = block[x / 3][y / 3][i] = 0;
}

void dfs(int x, int y)
{
    if (flag)
        return;
    if (x == 9)
    {
        for (int i = 0; i < 9; i++)
        {
            for (int j = 0; j < 9; j++)
            {
                cout << now[i][j];
            }
            puts("");
        }
        flag = true;
        return;
    }
    int nx = (y == 8 ? x + 1 : x);
    int ny = (y == 8 ? 0 : y + 1);
    if (now[x][y])
        dfs(nx, ny);
    else
    {
        for (int i = 1; i <= 9; i++)
        {
            if (check(x, y, i))
            {
                now[x][y] = i;
                dfs(nx, ny);
                back_trace(x, y, i);
            }
        }
    }
}
int main()
{
    cin >> n;
    char tmp;
    while (n--)
    {
        flag = false;
        for (int i = 0; i < 9; i++)
        {
            for (int j = 1; j <= 9; j++)
            {
                row[i][j] = col[i][j] = 0;
            }
        }
        for (int i = 0; i < 3; i++)
        {
            for (int j = 0; j < 3; j++)
            {
                for (int k = 1; k <= 9; k++)
                    block[i][j][k] = 0;
            }
        }
        for (int i = 0; i < 9; i++)
        {
            for (int j = 0; j < 9; j++)
            {
                cin >> tmp;
                now[i][j] = tmp - '0';
                if (now[i][j])
                {
                    row[i][now[i][j]] = 1;
                    col[j][now[i][j]] = 1;
                    block[i / 3][j / 3][now[i][j]] = 1;
                }
            }
        }
        dfs(0, 0);
    }
    return 0;
}