【DFS】【AOJ-61】Lake Counting

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
Line1: Two space-separated integers: N and M

Lines2..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

思路：
还是没啥好说的...深度优先搜索 不明白的翻数据结构图那部分

参考代码：

#include <stdio.h>
#include <string.h>
int visited[111][111],movex[8]={-1,-1,0,1,1,1,0,-1},movey[8]={0,-1,-1,-1,0,1,1,1},nx[100000],ny[100000];
char map[111][111];
void dfs(int i,int j);
int main()
{
int n,m;
scanf("%d%d",&n,&m);
int i,j;
memset(map,-1,sizeof(map));
memset(visited,0,sizeof(visited));
/*   for(i=1;i<=n;i++)
{
for(j=1;j<=m;j++)
{
char temp;
scanf("%c",&temp);
if(temp=='.')
map[i][j]=0;
else
map[i][j]=1;
}
getchar();
} */
memset(map,0,sizeof(map));
for(int i = 1;i <= n;i ++)
scanf("%s",&map[i][1]);
int num=0;
for(i=1;i<=n;i++)
{
for(j=1;j<=m;j++)
{
if(map[i][j]=='W'&&!visited[i][j])
{
num++;
dfs(i,j);
}
}
}
printf("%d\n",num);
return 0;
}

void dfs(int i,int j)
{
visited[i][j]=1;
int k;
for(k=0;k<8;k++)
{
if(map[i+movex[k]][j+movey[k]]=='W'&&!visited[i+movex[k]][j+movey[k]])
dfs(i+movex[k],j+movey[k]);
else
continue;
}
}