POJ DFS2386

最简单的DFS

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.

#include <stdio.h>
#include <stdlib.h>
#include <malloc.h>
#include <limits.h>
#include <ctype.h>
#include <string.h>
#include <string>
#include <math.h>
#include <algorithm>
#include <iostream>
#include <queue>
#include <stack>
#include <deque>
#include <vector>
#include <set>
//#include <map>
using namespace std;
#define MAXN 100 + 10
int n,m;

char map[MAXN][MAXN];
int vis[MAXN][MAXN];
int xx[8] = {1,1,0,-1,-1,-1,0,1};
int yy[8] = {0,1,1,1,0,-1,-1,-1};

void DFS(int x,int y){
    int i,j;
    for(i=0;i<8;i++){
        int dx = x+xx[i];
        int dy = y+yy[i];
        if(dx>=0 && dx<n && dy>=0 && dy<=m && map[dx][dy]=='W' && vis[dx][dy]==0){
            vis[dx][dy] = 1;
            DFS(dx,dy);
        }
    }
}

int main(){
    int i,j;

    while(~scanf("%d%d",&n,&m)){
        memset(vis,0,sizeof(vis));
        for(i=0;i<n;i++){
			//getchar();
            scanf("%s",map[i]);
        }
        int count = 0;
        for(i=0;i<n;i++){
            for(j=0;j<m;j++){
                if(map[i][j]=='W' && vis[i][j]==0){
                    vis[i][j] = 1;
                    DFS(i,j);
                    count++;
                }
            }
        }
        printf("%d\n",count);
    }

    return 0;
}