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

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

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\n", count);
}

return 0;
}