poj 2386 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

* 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

统计有多少组八连通W

#include <iostream>
#include <string.h>
#include <stdio.h>
using namespace std;
char map[105][105];
bool vis[105][105];
int dir[8][2]={{1,-1},{1,0},{1,1},{0,-1},{0,1},{-1,-1},{-1,0},{-1,1}};
int n,m,cnt;
bool judge(int x,int y)
{
if(!vis[x][y] && map[x][y]=='W'&& x>=0 && x<n && y>=0 && y<m)
{
return true;
}
return false;
}
void dfs(int x,int y)
{
vis[x][y]=true;
for (int i=0;i<8; i++)
{
if(judge(x+dir[i][0], y+dir[i][1]))
{
dfs(x+dir[i][0], y+dir[i][1]);
}
}
}
int main()
{
while (scanf("%d%d",&n,&m)!=EOF)
{
cnt=0;
for (int i=0; i<n; i++)
{
scanf("%s",map[i]);
}
memset(vis, false, sizeof(vis));
for (int i=0; i<n; i++)
{
for (int j=0; j<m; j++)
{
if(!vis[i][j] && map[i][j]=='W')
{
dfs(i, j);
cnt++;
}
}
}
printf("%d\n",cnt);
}
return 0;
}