poj 2386 DFS搜索基础

最初不会的地方：

1、  其实连样例开始的时候就没读懂……

2、  从哪里搜？

3、  搜的边界？

4、  把一片搜完了，然后呢？

5、  回溯吗？（这道题看到了回溯和DFS的区别）

标程的解决方法：

2、任何一个位置都可以，反复搜，所以从（0，0）位置起

3、搜的结果是把可以连为一片的W都标为1，W周围的.仍作为边界。

4、搜到所有的W都标为1的时候  搜索就结束了。（其实有这个原因：每个W必然属于一个区域，ans至少一个区域，当还有w没有被标记，就说明还有区域计数，当所有的W都被计数，就可以输出Ans）

另外，此题还然我学到几点：
一、int dir[2][8]={{-1,-1,-1,0,0,1,1,1},{-1,0,1,-1,1,-1,0,1}};

左边的{}是行坐标，右边是列坐标。

这样dfs的时候就不用写八个了……

void dfs(int i,int j)

{

   int k,a,b;

   for(k=0;k<8;k++)

    {

       a=dir[0][k]+i;

       b=dir[1][k]+j;

       if(a>=0&&a<=n&&b>=0&&b<=m&&!vis[a][b]&&map[a][b])

       {

           vis[a][b]=1;

           dfs(a,b);

       }

    }

}

二读入map直接记录为1，0

三、好好看本题的代码，体会回溯和DFs区别

贴代码：

 

 

Lake Counting

Time Limit: 1000MS	Memory Limit: 65536K
Total Submissions: 16081	Accepted: 8141

Description

Due to recentrains, water has pooled in various places in Farmer John's field, which isrepresented by a rectangle of N x M (1 <= N <= 100; 1 <= M <= 100)squares. Each square contains either water ('W') or dry land ('.'). Farmer Johnwould like to figure out how many ponds have formed in his field. A pond is aconnected set of squares with water in them, where a square is consideredadjacent to all eight of its neighbors. 

Given a diagram of Farmer John's field, determine how many ponds he has.

Input

* Line 1: Twospace-separated integers: N and M 

* Lines 2..N+1: M characters per line representing one row of Farmer John'sfield. Each character is either 'W' or '.'. The characters do not have spacesbetween them.

Output

* Line 1: Thenumber 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

OUTPUTDETAILS: 

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

Source

USACO2004 November

#include<cstdio>
#include<cstring>
using namespace std;
#define N 101

bool vis[N][N],map[N][N];
int fi,fj,m,n,count;
int dir[2][8]={{-1,-1,-1,0,0,1,1,1},{-1,0,1,-1,1,-1,0,1}};
int find()
{
    int i,j;
    for(i=0;i<n;i++)
        for(j=0;j<m;j++)
        {
            if(!vis[i][j]&&map[i][j])
            {
                fi=i;
                fj=j;
                return 1;
            }
        }
    return 0;
}

void dfs(int i,int j)
{
    int k,a,b;
    for(k=0;k<8;k++)
    {
        a=dir[0][k]+i;
        b=dir[1][k]+j;
        if(a>=0&&a<=n&&b>=0&&b<=m&&!vis[a][b]&&map[a][b])
        {
            vis[a][b]=1;
            dfs(a,b);
        }
    }
}

int main()
{
    int i,j;
    char c;

    while(scanf("%d%d",&n,&m)!=EOF)
    {
        memset(vis,0,sizeof(vis));
        memset(map,0,sizeof(map));
        count=0;
        for(i=0;i<n;i++)
            for(j=0;j<m;j++)
            {
               scanf("%c",&c);
               if(c=='W')map[i][j]=1;
               else if(c=='.')map[i][j]=0;
                    else if(c==' '||c=='\n'||c=='\0')j--;
            }
        while(find())
        {
            vis[fi][fj]=1;
            dfs(fi,fj);
            count++;
        }
        printf("%d\n",count);
    }

    return 0;
}