POJ 2386 Lake Counting（经典dfs）

Lake Counting
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 33538 Accepted: 16715
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

dfs经典题
题解：一次dfs后与初始的这个W连接的所有W都替换为‘.’,直到图中不存在W为止，总共进行的dfs的次数就是答案。

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>

using namespace std;
const int maxn = 105;
char field[maxn][maxn];
int n,m;
void dfs(int x,int y)//现在的位置
{
    field[x][y] = '.';//替换为.
    for(int dx = -1;dx<=1;dx++)//循环遍历八个方向
    {
        for(int dy = -1;dy<=1;dy++)
        {
            int nx = dx+x;
            int ny = dy+y;
            if(0<=nx&&nx<n&&0<=ny&&ny<m&&field[nx][ny] == 'W')//是否超出范围，是否有积水
                dfs(nx,ny);
        }
    }
    return;
}
int main()
{
    cin>>n>>m;
    char s;
    for(int i=0;i<n;i++)
    {
        for(int j=0;j<m;j++)
        {
            cin>>field[i][j];
        }
    }
    int cnt =0;
    for(int i=0;i<n;i++)
    {
        for(int j=0;j<m;j++)
        {
            if(field[i][j] == 'W')//从W开始遍历
                {
                    dfs(i,j);
                    cnt++;
                }
        }
    }
    cout<<cnt<<endl;
    return 0;
}