POJ2386 Lake Counting【DFS】
Lake Counting
| 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
3Hint
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
#include
#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