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。。。

 

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/************************************************ * 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; }