POJ2386 Lake Counting【DFS】

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 35139		Accepted: 17450

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

USACO 2004 November

问题链接：POJ2386 Lake Counting。

题意简述：给定m×n矩阵 (1 <= N <= 100; 1 <= M <= 100)，其中'W'代表水域，'.'代表陆地，问有几片湖。

本题可以使用深度优先搜索求解，用广度优先搜索也可以求解，差别不大。

问题分析：这个题与《UVa572 Oil Deposits》完全相同，程序改两个字符，改了一下结束条件就通过了。

程序说明：

程序中的有关内容说明如下：

1.方向数组　使用方向数组后，各个方向的试探的程序就会变得简洁了，用循环处理即可。

2.避免重复搜索　将搜索过的节点设置为'.'（陆地），可以避免重复搜索，能够简化程序逻辑。

3.设置边界　通过设置边界，可以免去矩阵（二维数组）的边界判断，简化了程序逻辑。

该问题与图遍历中寻找联通块问题基本上是同构的，算法思路一致。

每当找到一个水域，只需要计数加一，并且使用DFS算法把与其相邻的８个水域擦除即可（避免重复计数）。

参考链接：UVa572 Oil Deposits

AC的C语言程序如下：

/* POJ2386 Lake Counting */

#include <stdio.h>
#include <string.h>

#define DIRECTSIZE 8

struct direct {
    int drow;
    int dcol;
} direct[DIRECTSIZE] =
    {{0, -1}, {0, 1}, {-1, 0}, {1, 0}, {-1, -1}, {-1, 1}, {1, -1}, {1, 1}};

#define MAXN 100

char grid[MAXN+2][MAXN+2];

void dfs(int row, int col)
{
    int i;

    for(i=0; i<DIRECTSIZE; i++) {
        int nextrow = row + direct[i].drow;
        int nextcol = col + direct[i].dcol;

        if(grid[nextrow][nextcol] == 'W') {
            grid[nextrow][nextcol] = '.';

            dfs(nextrow, nextcol);
        }
    }
}

int main(void)
{
    int m, n, count, i, j;

    while(scanf("%d%d", &m, &n) != EOF) {
        // 清零：边界清零
        memset(grid, 0, sizeof(grid));

        // 读入数据
        for(i=1; i<=m; i++)
            scanf("%s", grid[i]+1);

        // 计数清零
        count = 0;

        // 深度优先搜索
        for(i=1; i<=m; i++)
            for(j=1; j<=n; j++)
                if(grid[i][j] == 'W') {
                    count++;
                    grid[i][j] = '.';
                    dfs(i, j);
                }

        // 输出结果
        printf("%d
", count);
    }

    return 0;
}