Lake Counting(DFS)

最新推荐文章于 2022-01-14 13:27:20 发布

Chien d'amis

最新推荐文章于 2022-01-14 13:27:20 发布

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分类专栏： ACM 文章标签： C++

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本文介绍了一种使用深度优先搜索(DFS)算法来解决湖泊计数问题的方法。在一个由水('W')和干地('.')组成的N*M网格中，湖泊被定义为相连水域的集合，每个方格与其八个相邻方格相连。通过遍历网格并应用DFS，可以确定湖泊的数量。文章提供了一个C++实现示例，展示了如何从给定的网格中计算湖泊总数。

Lake Counting(DFS)

Problem 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
在这里插入图片描述

Sample Output
3

运用DFS解决即可。

#include<iostream>
#include<cstdio>
using namespace std;
char visit[105][105];
int n, m;
int check(int x, int y)
{
	if (x >= 1 && x <=n && y >= 1 && y <=m)
		return 1;
	return 0;
}
void dfs(int x, int y)
{
	visit[x][y] = '.';
	for(int i=-1;i<=1;i++)
		for (int j = -1; j <= 1; j++)
		{
			if (check(x+i,y+j) && visit[x+i][y+j] == 'W')
				dfs(x+i, y+j);
		}
}
int main()
{
	int cnt=0;
	scanf("%d%d", &n, &m);
	for (int i = 1; i <=n; i++)
		for (int j = 1; j <=m; j++)
			cin >> visit[i][j];
	for (int i = 1; i <=n; i++)
		for (int j = 1; j <=m; j++)
		{
			if (visit[i][j] == 'W')
			{
				dfs(i, j);
				cnt++;
			}
		}
	printf("%d\n", cnt);
	return 0;
}