Lake Counting(DFS-深度优先搜索)

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.

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NxM(1<=N<=100;1<=M<=100)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.

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。

输出格式

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

一行：水坑的数量

输入输出样例

输入 #1复制

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

输出 #1复制

3

思路

 我们统计的时候需要注意去重，所以目标是将水坑去重。用dfs从第一出水洼开始判断旁边是否有连成一体的水坑，然后将连成一体的标记。每一次DFS标记好后，主函数循环继续下一步便找不到那些重复的水坑了，往后找到的都是独立的水坑。总共进行 DFS 的次数即为积水的次数。

#include <iostream>
using namespace std;
#define max 110
char a[max][max];
int ans=0, n, m;

void dfs(int x, int y)
{
	for (int i=-1; i<=1; i++)//八个方向遍历 
	for (int j=-1; j<=1; j++)
	{
		int dx = x+i;
		int dy = y+j;
		if (dx>=0 && dx<n && dy>=0 && dy<m && a[dx][dy]=='W')//如果没有越界而且是水坑 
		{
			a[dx][dy] = '.';//标记（去掉水坑标志，避免重复） 
			dfs(dx,dy);//继续探索它附近是否有连成一体的水坑 
		}
	}
	return;
}

int main()
{
	cin >> n >> m;
	
	for (int i=0; i<n; i++)
	for (int j=0; j<m; j++)
	{
		cin >> a[i][j];
	}
	
	for (int i=0; i<n; i++)
	for (int j=0; j<m; j++)
	{
		if (a[i][j] == 'W')
		{
			dfs(i,j);
			ans++;//水坑数 
		}		
	}
	
	cout << ans << endl;
}