POJ:2386 Lake Counting（dfs）
Lake Counting
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 40370 | Accepted: 20015 |
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.
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.
* 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.
There are three ponds: one in the upper left, one in the lower left,and one along the right side.
Source
题意：n行m列中，连在一起没有分开的W块有多少
从任意的W开始，不停地把邻接的部分用 . 代替。一次dfs后与初始的这个W连接的所有W就都被替换成了 . 。因此直到图中不再存在W为止，总共进行的dfs次数就是答案。
8个方向共对应了8种状态转移，每个格子作为dfs的参数至多被调用一次，复杂度为O（8*N*M）=O（N*M）
AC代码：
#include<cstdio>
#include<algorithm>
#include<iostream>
#include<cstring>
using namespace std;
int n,m;
const int maxn=1e3+10;
char ch[maxn][maxn];
void dfs(int x,int y)
{
ch[x][y]='.';
for(int dy=-1;dy<=1;dy++)
for(int dx=-1;dx<=1;dx++)
{
int nx=x+dx,ny=y+dy;
if(nx>=0&&nx<n&&ny>=0&&ny<m&&ch[nx][ny]=='W') dfs(nx,ny);//不断递归，把n*m区域内的W变成.
}
}
int main()
{
cin>>n>>m;
for(int i=0;i<n;i++)
for(int j=0;j<m;j++)
cin>>ch[i][j];
int sum=0;
for(int i=0;i<n;i++)
for(int j=0;j<m;j++)
{
if(ch[i][j]=='W')
{
dfs(i,j);
sum++;
}
}
cout<<sum<<endl;
return 0;
}