【题解】POJ 2386 LakeCounting(DFS)

最新推荐文章于 2020-12-22 11:36:28 发布

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最新推荐文章于 2020-12-22 11:36:28 发布

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分类专栏：挑战程序设计日常文章标签： dfs

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这篇博客介绍了如何使用深度优先搜索(DFS)解决POJ 2386 LakeCounting的问题。题目要求计算一块N x M的农场中，由'W'表示的积水区域形成的池塘数量。给定农场的布局后，通过DFS算法找出所有连通的积水区域并计数。博客内容包括问题描述、输入输出格式以及问题解决方案。

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POJ 2386 LakeCounting

原题

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.

https://vjudge.net/problem/POJ-2386

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

题意

因为下雨农场里有许多积水，连接在一起的积水为一个池塘，问一共有几片池塘。
’W’为有积水的土地，’.'为干燥的土地，这是一道基础的dfs题。
输入：第一行是农场的宽N和长M。(1 <= N, M <= 100)
接下来N+1行为每一行的土地情况，每行有M个字符分别为’W’，’.’。
输出：池塘的个数。
直接上代码

#include<iostream>
using namespace std;

char field[101][101];
int n, m, sum = 0;
void dfs(int x, int y)
{
	field[x][y] = '.';	//每找到一个'W'就把它标记为'.'
	for(int i = x - 1; i <= x + 1; i++)//该点周围一圈是否有积水
		for(int j = y - 1; j <= y + 1; j++)
			if(x >= 0 && x < n && y >= 0 && y < m && field[i][j] == 'W')
				dfs(i,j);
}
int main()
{
	cin >> n >> m;
	for(int i = 0; i < n; i++)
		for(int j = 0; j < m; j++)
			cin >> field[i][j];
	for(int i = 0; i < n; i++)
		for(int j = 0; j < m; j++)
			if(field[i][j] == 'W'){
				dfs(i,j);sum++;//每找到一个'W'都会把它自己以及与它相连的所有'W'都标记为'.'
			}					//所以直接用sum记录即可
	cout << sum << endl;
	return 0;
}