Lake Counting 搜索（水题）

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本文介绍了一种用于计算二维网格中湖泊数量的算法。通过深度优先搜索（DFS），遍历地图上的每个单元格，标记已访问过的水域并计算独立湖泊的数量。此算法适用于解决如雨水在农田中形成的水坑数量的问题。

									Lake Counting
				Time Limit: 1000MS		Memory Limit: 65536K
				Total Submissions: 47206		Accepted: 23209

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

3
Hint

OUTPUT DETAILS:

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

#include<bits/stdc++.h>
using namespace std;
int n,m;
char map1[101][101];
void dfs(int x,int y){
	int i,j,xx,yy;
	for(i = -1;i<=1;i++){
		for(j = -1;j<=1;j++){
			xx = x+i;
			yy = y+j;
			if (xx >= 0&&xx<n&&yy>=0&&yy<m&&map1[xx][yy] == 'W'){
				map1[xx][yy] = '.';
				dfs(xx,yy);
			}
		}
	}
}
int main(){
	int i,j,cnt = 0;
	cin>>n>>m;
	for(i = 0;i<n;i++){
		scanf("%s",map1[i]);
	}
	for(i = 0;i<n;i++){
		for(j = 0;j<m;j++){
			if(map1[i][j]=='W'){
				dfs(i,j);
				cnt++;
			}
		}
	}
	printf("%d\n",cnt);
}

水题一道没啥好讲的你们两个应该一看就懂
我去学优先队列或者结构体记忆搜索了要不然到时候凉凉可能最近2天出的解析少一些有些困难