[USACO10OCT] Lake Counting S

传送锚点：https://www.luogu.com.cn/problem/P1596

由于近期的降雨，雨水汇集在农民约翰的田地不同的地方。我们用一个 \(N\times M(1\leq N\leq 100, 1\leq M\leq 100)\) 的网格图表示。每个网格中有水（W） 或是旱地（.）。一个网格与其周围的八个网格相连，而一组相连的网格视为一个水坑。约翰想弄清楚他的田地已经形成了多少水坑。给出约翰田地的示意图，确定当中有多少水坑。

输入第 \(1\) 行：两个空格隔开的整数：\(N\) 和 \(M\)。

第 \(2\) 行到第 \(N+1\) 行：每行 \(M\) 个字符，每个字符是 W 或 .，它们表示网格图中的一排。字符之间没有空格。

输出一行，表示水坑的数量。

题目描述

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（池塘） 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.

输入格式

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.

输出格式

Line 1: The number of ponds in Farmer John's field.

样例 #1

样例输入 #1

10 12
W........WW.
.WWW.....WWW
....WW...WW.
.........WW.
.........W..
..W......W..
.W.W.....WW.
W.W.W.....W.
.W.W......W.
..W.......W.

样例输出 #1

3

提示

OUTPUT DETAILS: There are three ponds: one in the upper left, one in the lower left, and one along the right side.

思路

code

#include<iostream>
#include<vector>
#include<algorithm>
#include<cstring>
#include<cstdio>
using namespace std;
int n, m;//n为横向,m为纵向
const int maxn = 105;
int visited[maxn][maxn];//visited[i][j]为(坐标为(i,j))1代表访问过
char grid[maxn][maxn];//存储瓷砖内容
int res = 0;//统计水坑个数
int dx[8] = { -1,-1,-1,0,0,1,1,1 };
int dy[8] = { -1,0,1,-1,1,-1,0,1 };
void dfs(int x,int y) {//遍历的坐标
	for (int i = 0; i < 8; i++) {
		int nx = x + dx[i];//nx为下一步的横坐标
		int ny = y + dy[i];//ny为下一步的纵坐标
		if (nx < 0 || nx >= n || ny < 0 || ny >= m) continue;
		if (visited[nx][ny]) continue;
		if (grid[nx][ny] == '.') continue;
		visited[nx][ny] = 1;
		dfs(nx, ny);
	}

}
int main()
{
	cin >> n >> m;
	for (int i = 0; i < n; i++) {
		scanf("%s", grid[i]);
	}
	for (int i = 0; i < n; i++) {
		for (int j = 0; j < m; j++) {
			if (!visited[i][j] && grid[i][j] == 'W') {
				visited[i][j] = 1;
				dfs(i, j);
				res++;
			}
		}
	}
	cout << res;
	return 0;
}