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：

 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 using namespace std;

 char map[][];
 int mov[][]={,,,-,-,,,,,,-,-,,-,-,};
 int m,n;
 bool can(int x ,int y)//判断这个点能不能走
 {
     if(x<||x>m-||y<||y>n-||map[x][y]=='.')
     return false;
     return true;
 }

 void dfs(int x,int y)
 {
     int i,xx,yy;
     for(i=;i<;i++)
     {
         xx=x+mov[i][];
         yy=y+mov[i][];
         if(can(xx,yy))
         {
             map[xx][yy]='.';//走一个点就标记一下
             dfs(xx,yy);
         }
     }
 }
 int main()
 {
     int i,j;
     while(scanf("%d%d",&m,&n)!=EOF)
     {
         int sum=;
         for(i=;i<m;i++)
         scanf("%s",map[i]);
         for(i=;i<m;i++)
         {
             for(j=;j<n;j++)
             {
                 if(map[i][j]=='W')
                 {
                     map[i][j]='.';
                     dfs(i,j);//每次进入搜索函数就把这个点周围能走的点走完
                     sum++;
                 }
             }
         }
         printf("%d\n",sum);
     }
     return ;
 }

BFS版：

 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<queue>
 using namespace std;
 char map[][];
 int m,n;
 int mov[][]={,,,-,,,-,,,,-,-,,-,-,};
 struct node
 {
     int a,b;
 }ta,tb;//定义一个结构体用来存坐标
 bool can(node x)
 {
     if(x.a<||x.a>m-||x.b<||x.b>n-||map[x.a][x.b]=='.')
     return false;
     return true;
 }

 void bfs(int x,int y)
 {
     queue<node> q;
     ta.a=x;
     ta.b=y;
     q.push(ta);//入队
     while(!q.empty())//直到把队列能访问的都访问过
     {
         int i;
         ta=q.front();
         q.pop();
         for(i=;i<;i++)
         {
             tb.a=ta.a+mov[i][];
             tb.b=ta.b+mov[i][];
             if(can(tb))
             {
                 map[tb.a][tb.b]='.';
                 q.push(tb);//如果可以访问就入队
             }
         }
     }
 }

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