POJ DFS2386

最简单的DFS

Lake Counting

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 20516 Accepted: 10338

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 12W........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 + 10int 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;}