POJ No.2386（Lake Counting）

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

题意：

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.    W   .

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如上图：W可以连通八个方向。由题， 从W开始，先将W变为   .     ,往八个方向搜索是否有W。进行dfs搜索。dfs次数就是答案

复杂度O(n*m*8) = O(m*n)

代码:

#include <cstdio>
#include <iostream>
using namespace std;
const int maxn = 1e2 + 5;
char a[maxn][maxn];
int n, m;
void dfs(int x, int y) {
    a[x][y] = '.';
    for (int dx = - 1; dx <= 1; dx++)   //移动方向
        for (int dy = -1; dy <= 1; dy++) {
            int nx = x + dx, ny = y + dy;
            if (nx >= 0 && nx < n && ny >= 0 && ny < m && a[nx][ny] == 'W')
                dfs(nx, ny);
        }
    return;
}
int main()
{
    cin >> n >> m;
    for (int i = 0; i < n; i++)
        cin >> a[i];
    int res = 0;
    for (int i = 0; i < n; i++)
        for (int j = 0; j < m; j++)
            if (a[i][j] == 'W') {
                dfs(i, j);
                res++;
            }
    printf("%d\n", res);
    return 0;
}