Lake Counting

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本文详细解析了一种用于计算二维网格中湖泊数量的算法。该算法通过深度优先搜索（DFS）遍历网格，将相连的水域标记为同一湖泊，最终统计湖泊总数。适用于解决如美国计算机奥林匹克竞赛（USACO）等编程比赛中的类似问题。

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Lake Counting

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 48745		Accepted: 23941

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.

Source

USACO 2004 November

#include <iostream>
#include <cstdio>
#include <math.h>
#include <algorithm>
#include <string.h>
#include <string>

using namespace std;

const int N = 101,M = 101;

char map[N][M];
int n,m;

void dfs(int x,int y){
	
	map[x][y] = '.';
	
	for(int i=-1;i<=1;i++){
		for(int j=-1;j<=1;j++){
			int x1 = x + i;
			int y1 = y + j;
			if(x1>=0&&x1<N&&y>=0&&y<M&&map[x1][y1]=='W'){
				dfs(x1,y1);
			}
		}
	}
	
	return ;
}

int main(){
	
	ios::sync_with_stdio(false);
	
	while(cin>>n>>m){
		for(int i=0;i<n;i++){
		for(int j=0;j<m;j++){
			cin>>map[i][j];
		}
	}
	
	int res = 0;
	for(int i=0;i<n;i++){
		for(int j=0;j<m;j++){
			if(map[i][j] == 'W'){
				dfs(i,j);
				res++;
			}
		}
	}
	cout<<res<<endl;
	} 
	
	return 0;
}