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题目链接：https://www.luogu.org/problem/show?pid=1596
题目描述
Due to recent rains, water has pooled in various places in Farmer John‘s field, which is represented by a rectangle of N x M (1 <= N <= 100; 1 <= M <= 100) squares. Each square contains either water (‘W‘) or dry land (‘.‘). Farmer John would like to figure out how many ponds have formed in his field. A pond is a connected set of squares with water in them, where a square is considered adjacent to all eight of its neighbors. Given a diagram of Farmer John‘s field, determine how many ponds he has.
由于近期的降雨，雨水汇集在农民约翰的田地不同的地方。我们用一个NxM(1<=N<=100;1<=M<=100)网格图表示。每个网格中有水(‘W‘) 或是旱地(‘.‘)。一个网格与其周围的八个网格相连，而一组相连的网格视为一个水坑。约翰想弄清楚他的田地已经形成了多少水坑。给出约翰田地的示意图，确定当中有多少水坑。
输入输出格式
输入格式：
Line 1: Two space-separated integers: N and M *
Lines 2..N+1: M characters per line representing one row of Farmer
John‘s field. Each character is either ‘W‘ or ‘.‘. The characters do not
have spaces between them.
第1行：两个空格隔开的整数：N 和 M 第2行到第N+1行：每行M个字符，每个字符是‘W‘或‘.‘，它们表示网格图中的一排。字符之间没有空格。
输出格式：
Line 1: The number of ponds in Farmer John‘s field.
一行：水坑的数量
输入输出样例
输入样例#1：
10 12 W........WW. .WWW.....WWW ....WW...WW. .........WW. .........W.. ..W......W.. .W.W.....WW. W.W.W.....W. .W.W......W. ..W.......W.
输出样例#1：
3
分析：
深度优先搜索。
AC代码：
1 #include<iostream>
2 #include<cstdio>
3 #include<cmath>
4 #include<algorithm>
5
6 using namespace std;
7 char mp[110][110];
8 int m,n,ans = 0;
9 void dfs(int x,int y)
10 {
11 mp[x][y] = ‘.‘;
12 //通过将水坑标记为旱地，避免再次遍历
13 for(int i = -1;i <= 1;i ++)
14 for(int j = -1;j <= 1;j ++)
15 {
16 int nx = x + i;
17 int ny = y + j;
18 if(nx > 0 && nx <= n && ny > 0 && ny <= m && mp[nx][ny] == ‘W‘)
19 dfs(nx,ny);
20 }
21 return;
22 }
23
24 int main()
25 {
26 scanf("%d%d",&n,&m);
27 for(int i = 1;i <= n;i ++)
28 scanf("%s",mp[i] + 1);
29 for(int i = 1;i <= n;i ++)
30 for(int j = 1;j <= m;j ++)
31 {
32 if(mp[i][j] == ‘W‘)
33 dfs(i,j),ans ++;
34 }
35 printf("%d",ans);
36 return 0;
37 }