Lake Counting

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 41340		Accepted: 20504

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

//深度优先的典型题，大致意思是说找水坑

#include<iostream>
using namespace std;
char field[400][411];
int n, m;
void dfs(int i, int j)
{
    field[i][j] = '.';
    for(int dx=-1;dx<=1;dx++)
        for (int dy = -1; dy <= 1; dy++)
        {
            int k = i + dx, kk = j + dy;
            if (k >= 0 && k < n&& kk < m&&kk >= 0 && field[k][kk] == 'W')
                dfs(k, kk);
        }
    return;
}
int main()
{
    int mark = 0;
    cin >> n >> m;
    for (int i = 0; i < n; i++)
        for (int j = 0; j < m; j++)
            cin >> field[i][j];
    for(int i=0;i<n;i++)
        for (int j = 0; j < m; j++)
            if (field[i][j] == 'W')
            {
                dfs(i, j);
                mark++;
            }
    cout << mark << endl;
    return 0;
}

真正意义上的第一个深度题，代码不是很难理解，可是感觉掌握起来有点小麻烦；

还需多练习