POJ2386 Lake Counting 深搜

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：大小为N*M的菜园，雨后积起水，求出菜园*有多少水洼。

分析：从任意的‘W’开始进行DFS，不断把邻接的部分用'.'代替，1次DFS后与初始这个w连接的所有w就全都被替换成'.'，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。8个方向对应8个状态转移，每个格子作为DFS的参数最多调用一次，因此时间复杂度为O(8xNxM)=O(NM)。

#include<iostream>

using namespace std;
const int maxn=100;
char field[maxn][maxn]; 
int N,M;

void dfs(int x,int y)
{
	field[x][y]='.';
	
	for(int dx=-1;dx<=1;dx++)
		for(int dy=-1;dy<=1;dy++){
			int nx=x+dx,ny=y+dy;
			if(nx>=0&&nx<N&&ny>=0&&ny<M&&field[nx][ny]=='W'){
				dfs(nx,ny);
			}
		}
}
int main()
{
	int ans=0;
	cin>>N>>M;
	for(int i=0;i<N;i++){
		scanf("%s",field+i);
	}
	for(int i=0;i<N;i++)
		for(int j=0;j<M;j++){
			if(field[i][j]=='W'){
				dfs(i,j);
				ans++;
			}
		}
		cout<<ans<<endl;
		return 0;
}