DFS解决POJ 2386

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 <stdio.h>
char M[101][101];
int Used[101][101];
int direction[8][2]={{-1,-1},{-1,0},{-1,1},{0,-1},{0,1},{1,-1},{1,0},{1,1}};
int i,j,p;
int col,row;
int iCount=0;

void DFS(int x,int y)
{
    int k;
    for(k=0;k<8;k++)
    {
        int tx= x+direction[k][0];
        int ty= y+direction[k][1];
        if(tx >= 0 && tx < row && ty >= 0 && ty < col && M[tx][ty] == 'W' && Used[tx][ty]==0)
        {
                Used[tx][ty]=1;
                DFS(tx,ty);
        }
    }
}

int main()
{

    scanf("%d%d",&row,&col);
    for(i=0;i<row;i++)
        scanf("%s",M[i]);
    for(i=0;i<row;i++)
    {
        for(j=0;j<col;j++)
        {
            if(M[i][j]=='W' && Used[i][j] == 0)
            {
                    Used[i][j]=1;  //将其本身赋为已访问过(注意:这一点容易被忽略)
                    DFS(i,j);
                    iCount++;
            }
        }
    }
    printf("%d\n",iCount);
}