poj 2386 Lake Counting

Lake Counting

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 24578		Accepted: 12407

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.

Source

USACO 2004 November

#include<cstdio>
#include<cstring>
#include<iostream>
#include<stack>
#include<set>
#include<map>
#include<queue>
#include<algorithm>
using namespace std;
char Map[105][105];
int n,m;
void dfs(int x,int y){
    Map[x][y]='.';
    int i,j;
    for(i=-1;i<2;i++){
        int xx=x+i;
        for(j=-1;j<2;j++){
            int yy=y+j;
            if(xx>=0&&yy>=0&&xx<n&&yy<m&&Map[xx][yy]=='W'){
                dfs(xx,yy);
            }
        }
    }
}
int main(){
    //freopen("D:\INPUT.txt","r",stdin);
    scanf("%d %d",&n,&m);
    int i,j,k;
    for(i=0;i<n;i++){
        scanf("%s",Map[i]);
    }
    int res=0;
    for(i=0;i<n;i++){
        for(j=0;j<m;j++){
            if(Map[i][j]=='W'){
                dfs(i,j);
                res++;
            }
        }
    }
    printf("%d
",res);
    return 0;
}