题目传送门
1 /*
2 题意：问最少替换'*'为'.'，使得'.'连通的都是矩形
3 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
4 在2*2的方格里，若只有一个是'*'，那么它一定要被替换掉
5 */
6 #include <cstdio>
7 #include <iostream>
8 #include <algorithm>
9 #include <cstring>
10 #include <queue>
11 using namespace std;
12
13 const int MAXN = 2e3 + 10;
14 const int INF = 0x3f3f3f3f;
15 int n, m;
16 int dx[4][3] = {{1,0,1},{0,-1,-1},{-1,-1,0},{1,1,0}};
17 int dy[4][3] = {{0,1,1},{1,0,1},{0,-1,-1},{0,-1,-1}};
18 char s[MAXN][MAXN];
19
20 bool ok(int x, int y)
21 {
22 if (x < 0 || x >= n) return false;
23 if (y < 0 || y >= m) return false;
24
25 return true;
26 }
27
28 void BFS(void)
29 {
30 queue<pair<int, int> > Q;
31 for (int i=0; i<n; ++i)
32 {
33 for (int j=0; j<m; ++j)
34 {
35 if (s[i][j] == '.')
36 {
37 Q.push (make_pair (i, j));
38 }
39 }
40 }
41
42 while (!Q.empty ())
43 {
44 int x = Q.front ().first; int y = Q.front ().second;
45 Q.pop ();
46 for (int i=0; i<4; ++i)
47 {
48 int cnt = 0; int px, py; bool flag = true;
49 for (int j=0; j<3; ++j)
50 {
51 int tx = x + dx[i][j]; int ty = y + dy[i][j];
52 if (ok (tx, ty))
53 {
54 if (s[tx][ty] == '*')
55 {
56 cnt++; px = tx; py = ty;
57 }
58 }
59 else flag = false;
60 }
61 if (flag && cnt == 1)
62 {
63 s[px][py] = '.'; Q.push (make_pair (px, py));
64 }
65 }
66 }
67 }
68
69 int main(void) //Codeforces Round #297 (Div. 2) D. Arthur and Walls
70 {
71 while (scanf ("%d%d", &n, &m) == 2)
72 {
73 for (int i=0; i<n; ++i) scanf ("%s", s[i]);
74 BFS ();
75 for (int i=0; i<n; ++i) printf ("%s
", s[i]);
76 }
77
78 return 0;
79 }
80
81
82 /*
83 5 5
84 .*.*.
85 *****
86 .*.*.
87 *****
88 .*.*.
89 6 7
90 ***.*.*
91 ..*.*.*
92 *.*.*.*
93 *.*.*.*
94 ..*...*
95 *******
96 */