DFS POJ 2676 Sudoku

题目传送门

题意：数独问题，每行每列以及每块都有1~9的数字

分析：一个一个遍历会很慢。先将0的位子用vector存起来，然后用rflag[i][num] = 1 / 0表示在第i行数字num是否出现过，其他的类似，这样在用DFS就很快了，数据问题，反着搜索会更快。。。

/************************************************
* Author        :Running_Time
* Created Time  :2015/11/10 星期二 15:43:47
* File Name     :POJ_2676.cpp
 ************************************************/
 
#include <iostream>
#include <vector>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <cmath>
using namespace std;
 
#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 1e5 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const double EPS = 1e-10;
const double PI = acos (-1.0);
struct Pos  {
    int x, y;
    Pos () {}
    Pos (int x, int y) : x (x), y (y) {}
};
vector<Pos> blank;
int mp[11][11];
int rflag[11][11], cflag[11][11], bflag[11][11];
 
int get_id(int x, int y)    {
    int xx = x / 3;
    int yy = y / 3;
    return xx * 3 + yy;
}
 
void set_num(int x, int y, int num, int f) {
    rflag[x][num] = f;
    cflag[y][num] = f;
    bflag[get_id (x, y)][num] = f;
}
 
bool ok(int x, int y, int num)   {
    return !rflag[x][num] && !cflag[y][num] && !bflag[get_id (x, y)][num];
}
 
bool DFS(int cnt)   {
    if (cnt < 0)    return true;
    int x = blank[cnt].x, y = blank[cnt].y;
    for (int i=1; i<=9; ++i)    {
        if (!ok (x, y, i))  continue;
        mp[x][y] = i;
        set_num (x, y, i, 1);
        if (DFS (cnt - 1))  return true;
        set_num (x, y, i, 0);
    }
    return false;
}
 
int main(void)    {
    int T;  scanf ("%d", &T);
    while (T--) {
        for (int i=0; i<9; ++i)    {
            for (int j=0; j<9; ++j)    {
                scanf ("%1d", &mp[i][j]);
            }
        }
        blank.clear ();
        memset (rflag, 0, sizeof (rflag));
        memset (cflag, 0, sizeof (cflag));
        memset (bflag, 0, sizeof (bflag));
        for (int i=0; i<9; ++i)    {
            for (int j=0; j<9; ++j)    {
                if (mp[i][j] == 0)  blank.push_back (Pos (i, j));
                else    {
                    set_num (i, j, mp[i][j], 1);
                }
            }
        }
        if (DFS (blank.size () - 1))    {
            for (int i=0; i<9; ++i)    {
                for (int j=0; j<9; ++j)    {
                    printf ("%d", mp[i][j]);
                }
                puts ("");
            }
        }
        else    puts ("233");
    }
 
   //cout << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.\n";
 
    return 0;
}