POJ2386:Lake Counting(DFS)

Lake Counting

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 14776		Accepted: 7496

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.
Source

USACO 2004 November

MYCode:

#include<iostream>

#include<cstring>

#include<cstdio>

#define MAX 110

char map[MAX][MAX];

bool vis[MAX][MAX];

int dirt[8][2]={{1,0},{-1,0},{0,1},{0,-1},{1,1},{1,-1},{-1,1},{-1,-1}};

using namespace std;

void dfs(int x,int y)

{

vis[x][y]=1;

int i;

int tx,ty;

for(i=0;i<8;i++)

{

tx=x+dirt[i][0];

ty=y+dirt[i][1];

if(!vis[tx][ty] && map[tx][ty]=='W')

{

dfs(tx,ty);

}

}

}

int main()

{

int n,m;

while(scanf("%d%d",&n,&m)!=EOF)

{

int i,j;

memset(vis,0,sizeof(vis));

for(i=1;i<=n;i++)

{

for(j=1;j<=m;j++)

{

cin>>map[i][j];

}

}

int ans=0;

for(i=1;i<=n;i++)

{

for(j=1;j<=m;j++)

{

if(map[i][j]=='W'&&!vis[i][j])

{

ans++;

dfs(i,j);

}

}

}

printf("%d\n",ans);

}

}

//

DFS