BFS Codeforces Round #297 (Div. 2) D. Arthur and Walls

题目传送门

 /*
     题意：问最少替换'*'为'.'，使得'.'连通的都是矩形
     BFS：搜索想法很奇妙，先把'.'的入队，然后对于每个'.'八个方向寻找
         在2*2的方格里，若只有一个是'*'，那么它一定要被替换掉
 */
 #include <cstdio>
 #include <iostream>
 #include <algorithm>
 #include <cstring>
 #include <queue>
 using namespace std;

 const int MAXN = 2e3 + ;
 const int INF = 0x3f3f3f3f;
 int n, m;
 int dx[][] = {{,,},{,-,-},{-,-,},{,,}};
 int dy[][] = {{,,},{,,},{,-,-},{,-,-}};
 char s[MAXN][MAXN];

 bool ok(int x, int y)
 {
     if (x <  || x >= n)    return false;
     if (y <  || y >= m)    return false;

     return true;
 }

 void BFS(void)
 {
     queue<pair<int, int> > Q;
     for (int i=; i<n; ++i)
     {
         for (int j=; j<m; ++j)
         {
             if (s[i][j] == '.')
             {
                 Q.push (make_pair (i, j));
             }
         }
     }

     while (!Q.empty ())
     {
         int x = Q.front ().first;    int y = Q.front ().second;
         Q.pop ();
         for (int i=; i<; ++i)
         {
             int cnt = ;    int px, py;    bool flag = true;
             for (int j=; j<; ++j)
             {
                 int tx = x + dx[i][j];    int ty = y + dy[i][j];
                 if (ok (tx, ty))
                 {
                     if (s[tx][ty] == '*')
                     {
                         cnt++;    px = tx;    py = ty;
                     }
                 }
                 else    flag = false;
             }
             if (flag && cnt == )
             {
                 s[px][py] = '.';    Q.push (make_pair (px, py));
             }
         }
     }
 }

 int main(void)        //Codeforces Round #297 (Div. 2) D. Arthur and Walls
 {
     while (scanf ("%d%d", &n, &m) == )
     {
         for (int i=; i<n; ++i)    scanf ("%s", s[i]);
         BFS ();
         for (int i=; i<n; ++i)    printf ("%s\n", s[i]);
     }

     return ;
 }

 /*
 5 5
 .*.*.
 *****
 .*.*.
 *****
 .*.*.
 6 7
 ***.*.*
 ..*.*.*
 *.*.*.*
 *.*.*.*
 ..*...*
 *******
 */