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
解题思路:先把地图存到char数组中,然后遍历到一个W就dfs(i,j)把和它相邻的W全变成'.',然后ans++来记录池塘的个数
代码如下
#include<iostream>
using namespace std;
const int maxn=100+10;
int n,m;
char mp[maxn][maxn];
int dfs(int i,int j)
{
mp[i][j]='.';
for(int p=-1;p<=1;p++)
{
for(int q=-1;q<=1;q++)
{
int di=i+p,dj=j+q;
if(di>=0&&di<n&&dj>=0&&dj<m&&mp[di][dj]=='W')
dfs(di,dj);///将联通的'W'都变成'.'
}
}
}
void solve()
{
cin>>n>>m;
for(int i=0;i<n;i++)
{
for(int j=0;j<m;j++)
{
cin>>mp[i][j];
}
}
int ans=0;///记录池塘个数
for(int i=0;i<n;i++)
{
for(int j=0;j<m;j++)
{
if(mp[i][j]=='W')
{
dfs(i,j);
ans++;
}
}
}
cout<<ans<<endl;
}
int main()
{
ios::sync_with_stdio(0),cin.tie(0);
solve();
}