POJ 2386 Lake Counting (裸DFS 黑白图像)

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本文介绍了一个经典的湖泊计数问题，即在给定的矩形地图上寻找由'W'字符形成的独立水池数量。该问题可通过深度优先搜索（DFS）算法解决。

Lake Counting

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 21833		Accepted: 10993

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

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

题目链接：http://poj.org/problem?id=2386

题目大意：就是黑白图像问题

题目分析：DFS搜一下

#include <cstdio>
#include <cstring>
int const MAX = 205;
int dx[8] = {0, 0, 1, 1, 1, -1, -1, -1};
int dy[8] = {1, -1, 1, 0, -1, 1, 0, -1};
char map[MAX][MAX];
bool vis[MAX][MAX];
int n, m;

void DFS(int x, int y)
{
    vis[x][y] = true;
    for(int i = 0; i < 8; i++)
    {
        int xx = x + dx[i];
        int yy = y + dy[i];
        if(xx < 0 || yy < 0 || xx >= n|| yy >= m || vis[xx][yy] || map[xx][yy] != 'W')
            continue;
        DFS(xx, yy);
    }
}

int main()
{
    int ans = 0;
    scanf("%d %d", &n, &m);
    for(int i = 0; i < n; i++)
        scanf("%s", map[i]);
    memset(vis, false, sizeof(vis));
    for(int i = 0; i < n; i++)
    {
        for(int j = 0; j < m; j++)
        {
            if(map[i][j] == 'W' && !vis[i][j])
            {
                DFS(i, j);
                ans++;
            }
        }
    }
    printf("%d\n", ans);
}