POJ 2386 Lake Counting

最新推荐文章于 2019-09-06 19:25:48 发布

ZD.99

最新推荐文章于 2019-09-06 19:25:48 发布

阅读量139

点赞数

CC 4.0 BY-SA版权

分类专栏： POJ

本文链接：https://blog.youkuaiyun.com/qq_41072874/article/details/84786765

POJ 专栏收录该内容

2 篇文章

订阅专栏

本文介绍了一个使用深度优先搜索(DFS)算法解决的问题——计算二维矩阵中相连水洼的数量。该算法遍历矩阵，将每个水洼标记为已访问，并递归地检查其八个相邻单元格，直到所有相连的水洼都被访问过。通过这种方法，可以准确地计算出农田中形成的独立水洼总数。

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.

#include<iostream>
#include<cstdio>
using namespace std;
const int MAX_N = 100;
int n, m;
char field[MAX_N][MAX_N];
int dir[8][2] = { {1,0},{1,1},{0,1},{-1,1},{-1,0},{-1,-1},{0,-1},{1,-1} }; 
void dfs(int x, int y)
{
	field[x][y] = '.';
	for (int i = 0; i < 8; i++)
	{
		int nx = x + dir[i][0];
		int ny = y + dir[i][1];
		if (nx >= 0 && nx < n&&ny >= 0 && ny < m&&field[nx][ny] == 'W')
			dfs(nx, ny);
	}
}
int main()
{
	int res = 0;
	scanf("%d%d", &n, &m);
	for (int i = 0; i < n; i++)
		for (int j = 0; j < m; j++)
			scanf(" %c",&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);
				res++;
			}
	printf("%d\n", res);
	return 0;
}