dfs 积水块问题

Lake Counting 本题链接

题目描述

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

代码

#include
using namespace std;
int N,M;
char Map[110][110];
int f[8][2]={{0,-1},{0,1},{-1,0},{1,0},{-1,-1},{-1,1},{1,-1},{1,1}};///八个方向
void dfs(int x,int y)
{
    Map[x][y]='.';
    for(int i=0;i<8;i++)
    {
        int dx=x+f[i][0];
        int dy=y+f[i][1];
        if(dx>=0&&dx<N&&dy>=0&&dy<M&&Map[dx][dy]=='W')
        {
            dfs(dx,dy);
        }
    }
}
void solve()
{
    int ans=0;
    for(int i=0;i<N;i++)
    {
        for(int j=0;j<M;j++)
        {
            if(Map[i][j]=='W')
            {
                dfs(i,j);
                ans++;
            }
        }
    }
    cout<<ans<<endl;
}
int main()
{
    cin>>N>>M;
    for(int i=0;i<N;i++)
        for(int j=0;j<M;j++)
    {
        cin>>Map[i][j];
    }
    solve();
    return 0;
}