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
OUTPUT DETAILS:
There are three ponds: one in the upper left, one in the lower left,and one along the right side.
题目描述：N*M花园，求有少个水池，八连通的积水认为是连接在一起的。
思路:如果该点为水池，将该点’ W ‘替换为’ . ‘并深搜，每执行一次dsf，与’ W '相连的水池就全都被替换了，直到花园中不再有W，总共的dfs次数就是答案。
#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;
int n,m;
char a[105][105];
void dfs(int x,int y)
{
//将现在的位置替换为'.'
a[x][y]='.';
//循环遍历8个方向
for(int dx = -1; dx <= 1;dx++){
for(int dy=-1;dy<=1;dy++){
int nx=x+dx;
int ny=y+dy;
//判断是否在园子里以及是否有积水
if(nx>=0 && nx<n && ny>=0 && ny<m && a[nx][ny]=='W')dfs(nx,ny);
}
}
return ;
}
void solve(){
int res=0;
for(int i=0;i<n;i++){
for(int j=0;j<m;j++){
if( a[i][j] == 'W' ){
dfs(i,j); //将水池变为陆地
res++;
}
}
}
printf("%d\n",res);
}
int main()
{
scanf("%d%d",&n,&m);
getchar();
for(int i=0;i<n;i++){
for(int j=0;j<m;j++){
scanf("%c",&a[i][j]);
}
getchar(); //吸收换行符
}
solve();
return 0;
}