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
Hint
OUTPUT DETAILS:
There are three ponds: one in the upper left, one in the lower left,and one along the right side.
代码+题解
#include<cstdio>
#include<string>
#include<cstring>
using namespace std;
int n,m;
char arr[105][105];
void dfs(int i,int j)
{
if(i<0||i>=n||j<0||j>=m)return; //如果遇到非法位置,直接return。
arr[i][j]='.';
if(arr[i-1][j-1]=='W') //寻找八个方向是否有水。
dfs(i-1,j-1);
if(arr[i-1][j]=='W')
dfs(i-1,j);
if(arr[i-1][j+1]=='W')
dfs(i-1,j+1);
if(arr[i][j-1]=='W')
dfs(i,j-1);
if(arr[i][j+1]=='W')
dfs(i,j+1);
if(arr[i+1][j-1]=='W')
dfs(i+1,j-1);
if(arr[i+1][j]=='W')
dfs(i+1,j);
if(arr[i+1][j+1]=='W')
dfs(i+1,j+1);
return;
}
int main()
{
scanf("%d%d",&n,&m);
for(int i=0;i<n;i++)
{
getchar(); //输入字符串,注意换行。
for(int j=0;j<m;j++)
{
scanf("%c",&arr[i][j]);
}
}
int ans=0;
for(int i=0;i<n;i++) //寻找有水的地方
{
for(int j=0;j<m;j++)
{
if(arr[i][j]=='W')
{
dfs(i,j); //执行dfs函数。
ans++;
}
}
}
printf("%d\n",ans);
return 0;
}