算法

POJ2386总结

Lake Counting

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.

我的代码

#include<stdio.h>
int map[101][101];//记录地图数据,0代表地面,1代表水面 
int flag[101][101]={0};//记录此位置是否已经被搜索过 
int num=0;//记录水塘的总个数 
int n,m;
int bound(int x,int y)
//判断所给的坐标是否越界 
{
	if (x>=0&&x<n&&y>=0&&y<m)	return 1;
	else return 0;
}
int search(int x,int y)
//从指定的位置开始搜索直到找到整个池塘 
{
	//八个if分别是八个方向 
	if (bound(x-1,y-1))
	{
		if (!flag[x-1][y-1])
		{
			flag[x-1][y-1]=1;
			if (map[x-1][y-1])
			//如果是水面就沿着整个位置展开新的搜索,下同 
			{
				search(x-1,y-1);
			}
		}
	}
	if (bound(x-1,y))
	{
		if (!flag[x-1][y])
		{
			flag[x-1][y]=1;
			if (map[x-1][y])
			{
				search(x-1,y);
			}
		}
	}
	if (bound(x-1,y+1))
	{
		if (!flag[x-1][y+1])
		{
			flag[x-1][y+1]=1;
			if (map[x-1][y+1])
			{
				search(x-1,y+1);
			}
		}
	}
	if (bound(x,y-1))
	{
		if (!flag[x][y-1])
		{
			flag[x][y-1]=1;
			if (map[x][y-1])
			{
				search(x,y-1);
			}
		}
	}
	if (bound(x,y+1))
	{
		if (!flag[x][y+1])
		{
			flag[x][y+1]=1;
			if (map[x][y+1])
			{
				search(x,y+1);
			}
		}
	}
	if (bound(x+1,y-1))
	{
		if (!flag[x+1][y-1])
		{
			flag[x+1][y-1]=1;
			if (map[x+1][y-1])
			{
				search(x+1,y-1);
			}
		}
	}
	if (bound(x+1,y))
	{
		if (!flag[x+1][y])
		{
			flag[x+1][y]=1;
			if (map[x+1][y])
			{
				search(x+1,y);
			}
		}
	}
	if (bound(x+1,y+1))
	{
		if (!flag[x+1][y+1])
		{
			flag[x+1][y+1]=1;
			if (map[x+1][y+1])
			{
				search(x+1,y+1);
			}
		}
	}
}
int main()
{
	
	char c;
	scanf("%d%d",&n,&m);
	c=getchar();
	for (int i=0;i<n;i++)
	{
		for (int j=0;j<m;j++)
		{
			c=getchar();
			if (c=='W')	map[i][j]=1;
			else if (c=='.') map[i][j]=0;
		}
		c=getchar();
	}
	for (int i=0;i<n;i++)
	{
		for (int j=0;j<m;j++)
		{
			if (!flag[i][j])
			//如果没有访问过这个位置 
			{
				if (map[i][j])
				//如果此处是水面 
				{
					num++;//池塘数量++ 
					search(i,j);
				}
			}
		}
	}
	printf("%d",num);
	return 0;
} 

我的思路是遍历所有未访问过的点(flag[i][j]=0),如果是水面,就说明有一个新发现的水塘(水塘个数++),并且对其进行搜索,直到探索完整个这个池塘。一趟遍历下来之后就可以得到水塘的总个数。

反思

上网搜了一下其他人的做法,同样是dfs,但是明显思路比我清晰:

#include <stdio.h>
#define maxn 107

int n,m;
char g[maxn][maxn];
int dir[10][2] = {{-1,-1},{-1,0},{-1,1},{0,-1},{0,1},{1,0},{1,-1},{1,1}};

void dfs(int x,int y){    
    g[x][y] = '.';//此行代码意义重大,相当于将其置为已访问状态    
    for(int i = 0;i < 8;i++) {        
        int dx = x+dir[i][0];        
        int dy = y+dir[i][1];        
        if(dx>n||dx<1||dy<1||dy>m)            
            continue;        
        if(g[dx][dy] == '.')            
            continue;        
        dfs(dx,dy);    
    }
}
int main()
{    
    while(scanf("%d%d",&n,&m)!=EOF)    {        
        int ans = 0;        
        for(int i = 1;i <= n;i++)            
            scanf("%s",g[i]+1);  
        int cnt = 0;
        for(int i = 1;i <= n;i++)        
        for(int j = 1;j <= m;j++) {
            if(g[i][j] == 'W') {                
                ans++;                
                dfs(i,j);//把所有和该点相邻的W都变成.            
            }        
        }
        printf("%d\n",ans);    
    }    
    return 0;
}

与我不同的地方主要是,我是新开辟了空间来记录地图上未访问过的位置,而上面的解法是直接修改地图。对比起来我新开了一个数组,浪费了不必要的空间。

另外我的代码中表示八个方向的搜索也很愚蠢。在查询网络上其它人的代码之后,我找到了两种做法:

一是上面的代码使用的将八个方向存储在数组内,调用时遍历方向数组。

int dir[10][2] = {{-1,-1},{-1,0},{-1,1},{0,-1},{0,1},{1,0},{1,-1},{1,1}};

void dfs(int x,int y){    
    g[x][y] = '.';//此行代码意义重大,相当于将其置为已访问状态    
    for(int i = 0;i < 8;i++) {        
        int dx = x+dir[i][0];        
        int dy = y+dir[i][1];        
        if(dx>n||dx<1||dy<1||dy>m)            
            continue;        
        if(g[dx][dy] == '.')            
            continue;        
        dfs(dx,dy);    
    }
}

二是直接遍历方向(-1,0,1)。

for (int dx = -1; dx <= 1; dx++)//循环遍历连通的8个方向
    {
        for (int dy = -1; dy <= 1; dy++)
        {
            int nx = x + dx, ny = y + dy;//向x方向移动dx,向y方向移动dy,移动的结果为(nx,ny)
            if (0 <= nx&&nx <= N && 0 <= ny&&ny <= M&&field[nx][ny] == 'W')//判断(nx,ny)是不是在园子里,以及是否有积水
                dfs(nx, ny);
        }
    }

此外我的算法还有有一个问题:在主函数遍历过程中如果一个点是地面,即便已经访问过这个点,也没有能够及时地将其flag置为1,这样之后可能还是会访问这个点,可能会进行了不必要的遍历。

for (int i=0;i<n;i++)
	{
		for (int j=0;j<m;j++)
		{
			if (!flag[i][j])
			//如果没有访问过这个位置
			{
                                //此处应该添加一句 flag[i][j]=1;
				if (map[i][j])
				//如果此处是水面 
				{
					num++;//池塘数量++ 
					search(i,j);
				}
			}
		}
	}

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