POJ 2386 Lake Counting-优快云博客

本文链接：https://blog.youkuaiyun.com/cakaf/article/details/52094153

本文介绍了一道经典的深搜题目——湖计数（POJ2386）。任务是在给定的矩阵中计算由字符 'W' 表示的水域形成的独立湖泊数量。湖泊定义为相邻的 'W' 区域，相邻包括水平、垂直及对角方向。文章提供了完整的C语言实现代码。

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/*
POJ 2386:Lake Counting

总时间限制: 1000ms 内存限制: 65536kB
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
* 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
* Line 1: The number of ponds in Farmer John's field.
样例输入
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
3
*/

又是一道深搜题，大概意思就是找出地图中有多少个W，并且上下左右对角的W被认为是连接到一起的
上代码。。。

#include<stdio.h>
#include<string.h>
int m,n;
char s[110][120];
int count;
void DFS(int x,int y)
{
	s[x][y]='.';
	for(int dx=-1;dx<=1;dx++)
	{
		for(int dy=-1;dy<=1;dy++)
		{
			int nx=x+dx;
			int ny=y+dy;
			if(nx>=0&&nx<m&&ny>=0&&ny<n+1&&s[nx][ny]=='W')
				DFS(nx,ny);
		}
	}
}
int main()
{
	scanf("%d %d",&m,&n);
	for(int i=0;i<m;i++)
	{
		for(int j=0;j<n+1;j++)
		{
			scanf("%c",&s[i][j]);
		}
	}
	count=0;
	for(int i=0;i<m;i++)
	{
		for(int j=0;j<n+1;j++)
		{
			if(s[i][j]=='W')
			{
				DFS(i,j);
				count++;
			}
		}
	}
	printf("%d\n",count);
	return 0;
}