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 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

统计有多少组八连通W

#include <iostream>#include <string.h>#include <stdio.h>using namespace std;char map[105][105];bool vis[105][105];int dir[8][2]={{1,-1},{1,0},{1,1},{0,-1},{0,1},{-1,-1},{-1,0},{-1,1}};int n,m,cnt;bool judge(int x,int y){    if(!vis[x][y] && map[x][y]=='W'&& x>=0 && x<n && y>=0 && y<m)    {        return true;    }    return false;}void dfs(int x,int y){    vis[x][y]=true;    for (int i=0;i<8; i++)    {        if(judge(x+dir[i][0], y+dir[i][1]))        {            dfs(x+dir[i][0], y+dir[i][1]);        }    }}int main(){    while (scanf("%d%d",&n,&m)!=EOF)    {        cnt=0;        for (int i=0; i<n; i++)        {            scanf("%s",map[i]);        }        memset(vis, false, sizeof(vis));        for (int i=0; i<n; i++)        {            for (int j=0; j<m; j++)            {                if(!vis[i][j] && map[i][j]=='W')                {                    dfs(i, j);                    cnt++;                }            }        }        printf("%d\n",cnt);    }    return 0;}