D. Lakes in Berland

链接：

http://codeforces.com/problemset/problem/723/D

题意

给你一个n/*m的矩阵，然后你们有不少于k条湖泊，然后你需要使得一些湖泊变成陆地，使得湖泊的数量恰好等于k，问你至少填多少个水。

湖泊不与外界相邻。

题解：

直接dfs搜出每一条湖泊，然后放在优先队列里，从小到大去填满就好了。

代码：

 1 #include<iostream>
 2 #include<queue>
 3 #include<vector>
 4 #include<functional>
 5 using namespace std;
 6
 7 typedef pair<int, int> P;
 8 typedef pair<int, P> PP;
 9 const int maxn = 55;
10 char map[maxn][maxn];
11 int vis[maxn][maxn];
12 int dx[4] = { 1,0,-1,0 };
13 int dy[4] = { 0,1,0,-1 };
14 int n, m, sum, fg;
15 priority_queue<PP, vector<PP>, greater<PP> > lake;
16
17 void dfs(P s)
18 {
19     int x = s.first, y = s.second;
20     sum++;
21     vis[x][y] = 1;
22     if (x == 1 || x == n || y == 1 || y == m) fg = 0;
23     for (int i = 0; i < 4; i++)
24     {
25         int nx = x + dx[i];
26         int ny = y + dy[i];
27         if (nx < 1 || nx > n) continue;
28         if (ny < 1 || ny > m) continue;
29         if (map[nx][ny] == ‘*‘) continue;
30         if (vis[nx][ny]) continue;
31         dfs(P(nx, ny));
32     }
33 }
34
35 void dfs2(P s)
36 {
37     int x = s.first, y = s.second;
38     map[x][y] = ‘*‘;
39     for (int i = 0; i < 4; i++)
40     {
41         int nx = x + dx[i];
42         int ny = y + dy[i];
43         if (nx < 1 || nx > n)continue;
44         if (ny < 1 || ny > m)continue;
45         if (map[nx][ny] == ‘*‘)continue;
46         dfs2(P(nx, ny));
47     }
48 }
49
50 int main()
51 {
52     int k;
53     cin >> n >> m >> k;
54     for (int i = 1; i <= n; i++)
55         for (int j = 1; j <= m; j++)
56             cin >> map[i][j];
57     for (int i = 1; i <= n; i++)
58         for (int j = 1; j <= m; j++)
59             if (map[i][j] == ‘.‘ && vis[i][j] == 0) {
60                 fg = 1;
61                 sum = 0;
62                 dfs(P(i, j));
63                 if (fg) lake.push(PP(sum, P(i, j)));
64             }
65     int ans = 0;
66     while (lake.size() != k) {
67         ans += lake.top().first;
68         dfs2(lake.top().second);
69         lake.pop();
70     }
71     cout << ans << endl;
72     for (int i = 1; i <= n; i++) {
73         for (int j = 1; j <= m; j++)
74             cout << map[i][j];
75         cout << endl;
76     }
77     return 0;
78 }