题目：
Description
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.
Given a diagram of Farmer John's field, determine how many ponds he has.
Input
* 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.
* 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.
Output
* Line 1: The number of ponds in Farmer John's field.
题目大意： 找出池塘的个数，连在一起的W算一个池塘
解题思路：用种子填补法，利用双重循环和递归的方式（以某个点为中心向它的8个方向扫描）
如:
1 void dfs(int r,int c,int id) 2 { 3 if(r<0||r>=m||c<0||c>=n) //越界约束 4 return; 5 if(b[r][c]>0||a[r][c]!='W') //防重复查找和排除不是目标的点 6 return; 7 b[r][c]=id; 8 for(int dr=-1;dr<=1;dr++) 9 for(int dc=-1;dc<=1;dc++) 10 if(dr!=0||dc!=0) 11 dfs(r+dr,c+dc,id); 12 }
代码：
1 #include<iostream>
2 #include<cstdio>
3 #include<cstring>
4 using namespace std;
5 const int maxn=100+5;
6 char a[maxn][maxn];
7 int m,n,b[maxn][maxn];
8 void dfs(int r,int c,int id)
9 {
10 if(r<0||r>=m||c<0||c>=n)
11 return;
12 if(b[r][c]>0||a[r][c]!='W')
13 return;
14 b[r][c]=id;
15 for(int dr=-1;dr<=1;dr++)
16 for(int dc=-1;dc<=1;dc++)
17 if(dr!=0||dc!=0)
18 dfs(r+dr,c+dc,id);
19 }
20 int main()
21 {
22 int i,j;
23 while(scanf("%d %d",&m,&n)==2 &&m &&n)
24 {
25 for( i=0;i<m;i++)
26 scanf("%s",a[i]);
27 memset(b,0,sizeof(b));
28 int cnt=0;
29 for( i=0;i<m;i++)
30 for( j=0;j<n;j++)
31 if(b[i][j]==0&&a[i][j]=='W')
32 dfs(i,j,++cnt);
33 printf("%d
",cnt);
34 }
35 return 0;
36 }