2015 HUAS Summer Training#2 G

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.

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.

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 }