poj 2386 Lake Counting

Lake Counting

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 17917		Accepted: 9069

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

深搜

附代码

#include <iostream>
#include <cstring>
using namespace std;
int n,m,ans,book[1010][1010];
char a[1010][1010];
int dx[] = {-1,-1,-1,0,0,1,1,1};
int dy[] = {-1,0,1,-1,1,-1,0,1};
void dfs(int x,int y)
{
for(int i = 0;i<8;i++)
{
int xx = x + dx[i];
int yy = y + dy[i];

if(xx >= 0 && xx < n && yy >=0 && yy < m && a[xx][yy] == 'W' && book[xx][yy] == 0)
{
book[xx][yy] = 1;
dfs(xx,yy);
}
}
}
int main()
{
while(cin >> n >> m)
{
if(n == 0 && m == 0)
break;
memset(book,0,sizeof(book));
ans = 0;
for(int i = 0;i<n;i++)
for(int j = 0;j<m;j++)
cin >> a[i][j];
for(int i = 0;i<n;i++)
{
for(int j = 0;j<m;j++)
{
if(a[i][j] == 'W' && book[i][j] == 0)
{
ans++;
dfs(i,j);
}
}
}
cout<<ans<<endl;
}
}