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.
Sample Input
10 12 W........WW. .WWW.....WWW ....WW...WW. .........WW. .........W.. ..W......W.. .W.W.....WW. W.W.W.....W. .W.W......W. ..W.......W.
Sample Output
3
Hint
OUTPUT DETAILS:
There are three ponds: one in the upper left, one in the lower left,and one along the right side.
题意
求八连通块，如下：
. . .
. W .
. . .
题解
从任意的W开始，不停地把邻接的部分用‘.‘代替。1次DFS后与初始的这个W连接的所有W就都被替换成了‘.‘，
因此直到图中不再存在W为止，总共进行DFS的次数就是答案了。8个方向共对应了8种状态转移，每个格子作为DFS的参数至多被调用一次，所以复杂度为O（8*N*M）=O（N*M）。
1 #include<cstdio>
2 #include<algorithm>
3 #include<cstring>
4 #include<iostream>
5 using namespace std;
6 int n,m;char map[105][105];
7 void dfs(int i,int j){
8 map[i][j]=‘.‘;
9 for(int dx=-1;dx<=1;dx++)
10 for(int dy=-1;dy<=1;dy++){
11 int nx=i+dx,ny=j+dy;
12 if(nx>=0&&nx<n&&ny>=0&&ny<m&&map[nx][ny]==‘W‘) dfs(nx,ny);
13 }
14 return ;
15 }
16 int main(){
17 ios::sync_with_stdio(false);
18 cin>>n>>m;
19 for(int i=0;i<n;i++) for(int j=0;j<m;j++) cin>>map[i][j];
20 int ans=0;
21 for(int i=0;i<n;i++)
22 for(int j=0;j<m;j++){
23 if(map[i][j]==‘W‘){
24 dfs(i,j);
25 ans++;
26 }
27 }
28 printf("%d\n",ans);
29 }