POJ - 2386 Lake Counting DFS

湖泊计数算法

最新推荐文章于 2021-06-18 17:10:57 发布

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本文介绍了一个湖泊计数算法，该算法使用深度优先搜索（DFS）来遍历地图上的每个单元格，确定水区域是否构成独立的湖泊。湖泊定义为一个连通的水域集合，每个水域可以通过其八个相邻方向连接到其它水域。文章提供了完整的C++实现代码。

Lake Counting

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 36120		Accepted: 17934

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

AC代码：

#include<cstdio>
char s[110][110];
int n, m;
void dfs(int x, int y){
	int next[8][2] = {{-1,0},{-1,1},{0,1},{1,1},{1,0},{1,-1},{0,-1},{-1,-1}};
	s[x][y] = '.';
	for(int k=0;k<=7;k++){
		int tx, ty;		
		tx = x + next[k][0];
		ty = y + next[k][1];
		if(tx<1 || ty<1 || tx>n || ty>m) continue; 
		if(s[tx][ty] == 'W'){
			dfs(tx, ty);
		}
	}
}
int main(){
	
	scanf("%d%d", &n, &m);
	getchar();
	for(int i = 1; i <= n; i++){
		for(int j = 1; j <= m; j++){
			scanf("%c", &s[i][j]);
		}
		getchar();
	}
	int sum = 0;
	for(int i = 1; i <= n; i++){
		for(int j = 1;j <= m; j++){
			if(s[i][j] == 'W'){
				sum++;
				dfs(i,j);
			}
		}
	} 
	printf("%d",sum);
	return 0;
}