POJ 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.

题目链接

http://poj.org/problem?id=2386

题目翻译

有一个大小为 N×M 的园子，雨后积起了水。八连通的积水被认为是连通在一起的（W表示积水，积水的上下左右和左上右上左下右下这八个方向如果有积水，则认为它们是连通的）。请求出园子里总共有多少水洼？
N，M<=100

题目分析

深度优先搜索即可。
遍历每一个点，如果值为W，则以这个点为起点开始深搜，答案 ans++。
在深搜的过程中，首先将当前点的值更改为.，然后遍历当前点的八个方向上的点，如果值为W，则以其为起点进行深搜。
具体看代码吧。

AC代码

#include <iostream>
using namespace std;

int n, m;
char num[105][105];

void dfs(int x, int y)
{
    num[x][y] = '.';
    int d[] = {0, -1, 1}, i, j;
    for (i = 0; i < 3; i++)
    {
        for (j = 0; j < 3; j++)
        {
            int x1 = x + d[i], y1 = y + d[j];
            if (x1 < n && x1 >= 0 && y1 < m && y1 >= 0 && num[x1][y1] == 'W')
            {
                dfs(x1, y1);
            }
        }
    }
}

int main()
{
    int i, j, ans = 0;
    cin >> n >> m;
    for (i = 0; i < n; i++)
        cin >> num[i];
    for (i = 0; i < n; i++)
    {
        for (j = 0; j < m; j++)
        {
            if (num[i][j] == 'W')
            {
                dfs(i, j);
                ans++;
            }
        }
    }
    cout << ans << endl;
    return 0;
}