POJ 2386 Lake Counting(dfs or bfs)

Lake Counting

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 24999		Accepted: 12619

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

题意：图给出的是一片农场，'W'表示水， '.' 表示土地。问这片农场有多少块池塘。

dfs解法：从任意的W开始，不停地把邻接的部分用‘.’代替。1次dfs后与初始的W连接的所有W就都被替换成了‘.’ ，

直到图中不再存在W为止，总共进行的dfs次数就是答案。

代码如下：

#include<cstdio>
int n,m;
char map[110][110];
int dir[8][2]={{-1,-1},{-1,0},{-1,1},{0,-1},{0,1},{1,-1},{1,0},{1,1}};

void dfs(int x,int y)
{
	int i,nowx,nowy;
	map[x][y]='.';//将W替换成‘.’ 
	for(i=0;i<8;++i)//遍历它的旁边8个方向 
	{
		nowx=x+dir[i][0];
		nowy=y+dir[i][1];
		if(nowx>=0&&nowx<n&&nowy>=0&&nowy<m&&map[nowx][nowy]=='W')
			dfs(nowx,nowy);
	}
}

int main()
{
	int i,j,cnt;
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		for(i=0;i<n;++i)
			scanf("%s",map[i]);
		cnt=0;
		for(i=0;i<n;++i)
		{
			for(j=0;j<m;++j)
			{
				if(map[i][j]=='W')
				{
					dfs(i,j);
					cnt++;
				}
			}
		}
		printf("%d\n",cnt);
	}
	return 0;
} 

bfs解法，解题思想与dfs一致，时间和内存也差不多

代码如下：

#include<cstdio>
#include<queue>
using namespace std;
#define maxn 110
char map[maxn][maxn];
int n,m;
int dir[8][2]={{-1,-1},{-1,0},{-1,1},{0,-1},{0,1},{1,-1},{1,0},{1,1}};
struct node
{
	int x,y;
}a,temp;

void bfs(int x,int y)
{
	int i;
	queue<node>q;
	a.x=x;
	a.y=y;
	q.push(a);
	map[x][y]='.';
	while(!q.empty())
	{
		a=q.front();
		q.pop();
		for(i=0;i<8;++i)
		{
			temp.x=a.x+dir[i][0];
			temp.y=a.y+dir[i][1];
			if(temp.x>=0&&temp.x<n&&temp.y>=0&&temp.y<m&&map[temp.x][temp.y]=='W')
			{
				q.push(temp);
				map[temp.x][temp.y]='.';
			}
		} 
	}
}

int main()
{
	int i,j,cnt;
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		for(i=0;i<n;++i)
			scanf("%s",map[i]);
		cnt=0;
		for(i=0;i<n;++i)
		{
			for(j=0;j<m;++j)
			{
				if(map[i][j]=='W')
				{
					bfs(i,j);
					cnt++;
				}
			}
		}
		printf("%d\n",cnt);
	}
	return 0;
}