POJ_2386: Lake Counting

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

备注：简单的dfs数连通分量个数。

#include<stdio.h>

#define LAND -1
#define WATER 1

int N,M;
int graph[101][101];
int visited[101][101];

void dfs(int startA,int startB)
{
visited[startA][startB]=true;

if(startA-1>=0 && !visited[startA-1][startB] && graph[startA-1][startB]==WATER)
dfs(startA-1,startB);
if(startA+1<N && !visited[startA+1][startB] && graph[startA+1][startB]==WATER)
dfs(startA+1,startB);
if(startB-1>=0 && !visited[startA][startB-1] && graph[startA][startB-1]==WATER)
dfs(startA,startB-1);
if(startB+1<M && !visited[startA][startB+1] && graph[startA][startB+1]==WATER)
dfs(startA,startB+1);
if(startA-1>=0 && startB-1>=0 && !visited[startA-1][startB-1] && graph[startA-1][startB-1]==WATER)
dfs(startA-1,startB-1);
if(startA-1>=0 && startB+1<M && !visited[startA-1][startB+1] && graph[startA-1][startB+1]==WATER)
dfs(startA-1,startB+1);
if(startA+1<N && startB-1>=0 && !visited[startA+1][startB-1] && graph[startA+1][startB-1]==WATER)
dfs(startA+1,startB-1);
if(startA+1<N && startB+1<M && !visited[startA+1][startB+1] && graph[startA+1][startB+1]==WATER)
dfs(startA+1,startB+1);
}

int main()
{

scanf("%d %d",&N,&M);
getchar();

for(int i=0;i<N;i++)
{
for(int j=0;j<M;j++)
{
char c;
c=getchar();
if(c=='.')
graph[i][j]=LAND;
else if(c=='W')
graph[i][j]=WATER;
visited[i][j]=false;
}
getchar();
}
//count connected components using dfs
int count=0;
for(int i=0;i<N;i++)
for(int j=0;j<M;j++)
{
if(!visited[i][j] && graph[i][j]==WATER)
{
dfs(i,j);
count++;
}
}

printf("%d\n",count);
return 0;
}