Lake Counting--poj2386

Lake Counting

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 23950		Accepted: 12099

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



这个题可以用深搜也可以用广搜，我就是从这个题，明白了两种搜索方式的不同

大家来体会一下吧！





DFS版：

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<algorithm>
 4 using namespace std;
 5 
 6 char map[101][101];
 7 int mov[8][2]={0,1,0,-1,-1,0,1,0,1,1,-1,-1,1,-1,-1,1};
 8 int m,n;
 9 bool can(int x ,int y)//判断这个点能不能走
10 {
11     if(x<0||x>m-1||y<0||y>n-1||map[x][y]=='.')
12     return false;
13     return true;
14 }
15 
16 void dfs(int x,int y)
17 {
18     int i,xx,yy;
19     for(i=0;i<8;i++)
20     {
21         xx=x+mov[i][0];
22         yy=y+mov[i][1];
23         if(can(xx,yy))
24         {
25             map[xx][yy]='.';//走一个点就标记一下
26             dfs(xx,yy);
27         }
28     }
29 }
30 int main()
31 {
32     int i,j;
33     while(scanf("%d%d",&m,&n)!=EOF)
34     {
35         int sum=0;
36         for(i=0;i<m;i++)
37         scanf("%s",map[i]);
38         for(i=0;i<m;i++)
39         {
40             for(j=0;j<n;j++)
41             {
42                 if(map[i][j]=='W')
43                 {
44                     map[i][j]='.';
45                     dfs(i,j);//每次进入搜索函数就把这个点周围能走的点走完
46                     sum++;
47                 }
48             }
49         }
50         printf("%d\n",sum);
51     }
52     return 0;
53 }

 

BFS版：

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<algorithm>
 4 #include<queue>
 5 using namespace std;
 6 char map[101][101];
 7 int m,n;
 8 int mov[8][2]={0,1,0,-1,1,0,-1,0,1,1,-1,-1,1,-1,-1,1};
 9 struct node
10 {
11     int a,b;
12 }ta,tb;//定义一个结构体用来存坐标
13 bool can(node x)
14 {
15     if(x.a<0||x.a>m-1||x.b<0||x.b>n-1||map[x.a][x.b]=='.')
16     return false;
17     return true;
18 }
19 
20 
21 void bfs(int x,int y)
22 {
23     queue<node> q;
24     ta.a=x;
25     ta.b=y;
26     q.push(ta);//入队
27     while(!q.empty())//直到把队列能访问的都访问过
28     {
29         int i;
30         ta=q.front();
31         q.pop();
32         for(i=0;i<8;i++)
33         {
34             tb.a=ta.a+mov[i][0];
35             tb.b=ta.b+mov[i][1];
36             if(can(tb))
37             {
38                 map[tb.a][tb.b]='.';
39                 q.push(tb);//如果可以访问就入队
40             }
41         }
42     }
43 }
44 
45 
46 int main()
47 {
48     int i,j;
49     while(scanf("%d%d",&m,&n)!=EOF)
50     {
51         int sum=0;
52         for(i=0;i<m;i++)
53         scanf("%s",map[i]);
54         for(i=0;i<m;i++)
55         for(j=0;j<n;j++)
56         {
57             if(map[i][j]=='W')
58             {
59                 map[i][j]='.';
60                 bfs(i,j);
61                 sum++;
62             }
63         }
64         printf("%d\n",sum);
65     }
66     return 0;
67 }