搜索_Lake Counting

source

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

DFS Solution

#include
#include
#include
#include
using namespace std;
 char garden[102][102];
int n,m;
void dfs(int i,int j)
{
    if(i<1||j<1||i>n||j>m) return ;
    garden[i][j]='.';
    for(int dx=i-1;dx<=i+1;dx++)
        for(int dy=j-1;dy<=j+1;dy++)
        {
              if(garden[dx][dy]=='W')
              {
                  dfs(dx,dy);
              }

        }


}
int main()
{

    int sum=0;
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++)
        for(int j=1;j<=m;j++)
        cin>>garden[i][j];
    for(int i=1;i<=n;i++)
        for(int j=1;j<=m;j++)
        {
            if(garden[i][j]=='W')
            {
                dfs(i,j);
                sum++;
            }

        }
        printf("%d",sum);

}

BFS Solution

#include
#include
using namespace std;
int main()
{
    char maze[102][102];
    int n,m,sum=0;
    pair p;
    queue > myqueue;
    scanf("%d%d",&n,&m);
    for(int i=0;i (i,j));
                while(!myqueue.empty())
                {
                    p=myqueue.front();
                    myqueue.pop();
                    for(int dx=p.first-1;dx<=p.first+1;dx++)
                        for(int dy=p.second-1;dy<=p.second+1;dy++)
                        {
                            if(dx>-1&&dy>-1&&dx (dx,dy));

                            }
                        }
                }
                sum++;
            }
        }
    }
    printf("%d",sum);
}