Lake Counting(深搜妙解)

Lake Counting

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 16840		Accepted: 8548

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

 

 

/*
思路：从任一W开始，不停把邻接部分用'.'代替。调用n次dfs就有n个积水。
*/
#include<iostream>
#define MAX 101

using namespace std;

int N,M;
char field[MAX][MAX];

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, ny = y + dy;

        if(0<=nx && nx<N && 0<=ny && ny<=M && field[nx][ny]=='W') dfs(nx, ny);
    }
}


int main()
{
    int ans=0,i,j;

    cin>>N>>M;

    for(i=0;i<N;i++)
        for(j=0;j<M;j++)
            cin>>field[i][j];

    for(i=0;i<N;i++)for(j=0;j<M;j++)
        if(field[i][j] == 'W')
        {dfs(i, j); ans++;}

    cout<<ans<<endl;

    /*system("pause");*/
    return 0;
}