lake counting---dfs(深度搜索算法）

问题描述

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.

输入描述
* 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.

输出描述
* Line 1: The number of ponds in Farmer John's field.

样例输入

10 12
W........WW.
.WWW.....WWW
....WW...WW.
.........WW.
.........W..
..W......W..
.W.W.....WW.
W.W.W.....W.
.W.W......W.
..W.......W.

样例输出

3

题目分析：

dfs实际上就是一直顺着一个方向不断地搜索直到搜索到头为止。并且在搜索的过程中，要进行标记，防止走过得路重复走。

以第一个点'W'点为例，以这个为节点分别向八个方向搜索，遇到新的'W'点时，在以这个新的'W'点为节点分别向八个方向搜索，不断递归，把一块区域搜索完。(0,0)--(1,1)--(1,2)--(1,3)--(2,4)--(2,5)一块区域搜索完毕。

代码如下：

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
int dx[8]={0,0,1,1,-1,-1,1,-1};
int dy[8]={1,-1,1,-1,-1,1,0,0};
char mp[102][102];
int node[102][102];///标记数组，走过的路就不要在走了，否则会陷入死循环。
int num,n,m;
void dfs(int x,int y)
{
int i,x1,y1;
if(x<0||y<0||x>=n||y>=m) return;
if(mp[x][y]=='.') return;
node[x][y]=1;
for(i=0;i<8;i++)///对八个方向进行搜索。
{
x1=x+dx[i];
y1=y+dy[i];
if(mp[x1][y1]=='W'&&node[x1][y1]==0) dfs(x1,y1);
}
}
int main()
{
int i,j;
///scanf("%d%d",&n,&m);
while(scanf("%d%d",&n,&m)!=EOF)
{
num=0;
memset(node,0,sizeof(node));///对标记数组进行初始化。
for(i=0; i<n; i++)
{
scanf("%s",mp[i]);
}
///深度搜索，对数组进行遍历。
for(i=0; i<n; i++)
{
for(j=0; j<m; j++)
{
if(node[i][j]==0&&mp[i][j]=='W')
{
dfs(i,j);
num++;///用来表示有没有连通的区域
}
}
}
printf("%d\n",num);
}
return 0;
}