# poj2386 Lake Counting（简单DFS）

u012860063 分享于 2014-07-15

Lake Counting Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 20782   Accepted: 10473 Description Due to recent rains, water has pooled i

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`

```#include <iostream>
#include <algorithm>
using namespace std;
#include <cstring>
#define TM 100+17
int N, M;
char map[TM][TM];
bool vis[TM][TM];
int xx={0,1,1,1,0,-1,-1,-1};
int yy={1,1,0,-1,-1,-1,0,1};
void DFS(int x, int y)
{
vis[x][y] = true;
for(int i = 0; i < 8; i++)
{
int dx = x+xx[i];
int dy = y+yy[i];
if(dx>=0&&dx<N&&dy>=0&&dy<M&&!vis[dx][dy]&&map[dx][dy] == 'W')
{
vis[dx][dy] = true;
DFS(dx,dy);
}
}
}
int main()
{
int i, j;
while(cin>>N>>M)
{
int count = 0;
memset(vis,false,sizeof(vis));
for(i = 0; i< N; i++)
{
cin>>map[i];
}
for(i = 0; i < N; i++)
{
for(j = 0; j < M; j++)
{
if(map[i][j] == 'W' && !vis[i][j])
{
count++;
DFS(i,j);
}
}
}
cout<<count<<endl;
}
return 0;
}```

Lake Counting Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 24999 Accepted: 12619 Description Due to recent rains, water has pooled in va  ×
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