POJ-2386 Lake Counting

Lake Counting

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 56258		Accepted: 27405

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 <iostream>
#include <cstdio>

using namespace std;

const int max_N=100+5;
const int max_M=100+5;

int N,M;
char field[max_N][max_M];

void dfs(int x,int y)
{
    field[x][y]='.';
    for(int dx=-1;dx<=1;++dx)
    {
        for(int dy=-1;dy<=1;++dy)
        {
            int nx=x+dx;
            int ny=y+dy;
            if(nx>=0 && nx<N && ny>=0 && ny<M && field[nx][ny]=='W')
            {
                dfs(nx,ny);
            }
        }
    }
}


void solve()
{
    int res=0;
    for(int i=0;i<N;++i)
    {
        for(int j=0;j<M;++j)
        {
            if(field[i][j]=='W')
            {
                dfs(i,j);
                ++res;
            }
        }
    }
    printf("%d
",res);
}

int main()
{
    scanf("%d %d",&N,&M);
    for(int i=0;i<N;++i)
    {
        scanf("%s",field[i]);
    }
    solve();
    return 0;
}