洛谷P1596水坑计数

[USACO10OCT] Lake Counting S

题目链接

题面翻译

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 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.

代码：

#include<iostream>
using namespace std;
int n, m,ans;
char g[1050][1050];
//int dx[8] = {-1,-1,-1,0,0,1,1,1};
//int dy[8] = {-1,0,1,-1,1,-1,0,1};
int dx[8] = { -1,-1,-1,0,1,1,1,0 };
int dy[8] = { -1,0,1,1,1,0,-1,-1 };
void dfs(int x, int y) {
	g[x][y] = '.';
	for (int i = 0; i < 8; i++) {
		int a = x + dx[i];
		int b = y + dy[i];
		if (a < 0 || a >= n || b < 0 || b >= m) continue;
		if (g[a][b] == '.') continue;
		dfs(a, b);
	}
}
int main()
{
	cin >> n >> m;
	for (int i = 0; i < n; i++) {
		cin >> g[i];
	}
	//枚举每一个水坑，只要是水坑，我就进去深搜，并且对每个点做变性手术
	for (int i = 0; i < n; i++) {
		for (int j = 0; j < m; j++) {
			if (g[i][j] == 'W') {
				ans++; dfs(i, j);
			}
		}
	}
	cout << ans;
	return 0;
}

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• 注意判重，避免死循环
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