POJ-2386 Lake Counting （深搜寻找）

Lake Counting

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 44144		Accepted: 21808

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

#include <cstdio>
#include <iostream>
using namespace std;
#define MAXN 130
int next[8][2]={{0,-1},{1,-1},{1,0},{1,1},{0,1},{-1,1},{-1,0},{-1,-1}};  //不自己写，这个细节都忘记了，吐血。
char str[MAXN][MAXN];
int n,m;
void dfs(int x,int y)
{
    str[x][y]='.';
    int i;
    for(i=0;i<8;i++)
    {
        int xn=x+next[i][0];
        int yn=y+next[i][1];
        if(xn<=0||xn>n||yn<=0||yn>m)
        continue;
        if(str[xn][yn]=='W')
        dfs(xn,yn);
    }
}
int main()
{
    int res=0;
    int i,j;
    scanf("%d%d",&n,&m);
    for(i=1;i<=n;i++)
    {
      getchar();
      for(j=1;j<=m;j++)
      scanf("%c",&str[i][j]);
    }
    for(i=1;i<=n;i++)
    for(j=1;j<=m;j++)
    {
        if(str[i][j]=='W')
        {
             dfs(i,j);
             res++;
        }
    }
    printf("%d\n",res);
    return 0;
}