poj2386(简单的dfs/bfs）

题目链接：http://poj.org/problem?id=2386

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.

题意：有一个大小为N*M的园子，雨后起了水。八连通的积水都被认为是连接在一起的。请求出园子里总共有多少水洼？（八连通指的是下图中相对W的*部分）

***

*W*

***

解题思路：从任意的W开始，不停把邻接的部分用'.'代替。1次dfs后与初始的这个W连接的所有W就都被替换成了'.'，因此知道图中不在出现'W'为止，总共进行dfs的次数就是最后的答案了，8个方向对应8种状态转移，每个格子作为dfs的参数之多被调用一次，所以复杂度为O(8*n*m)=O(n*m)。

附上代码：

#include<iostream>
#include<cstdio>
using namespace std;
char map[105][105];
int n,m;
int dir[8][2]={{-1,-1},{-1,0},{-1,1},{0,-1},{0,1},{1,-1},{1,0},{1,1}};
//现在位置为(x,y) 
void dfs(int x,int y)
{
    //将现在所在位置替换为. 
    map[x][y]='.';
    //循环遍历移动的8个方向 
    for(int i=0;i<8;i++)
    {
        int dx=x+dir[i][0];  
        int dy=y+dir[i][1];
        //判断(dx,dy)内是否又水 
        if(dx>=0&&dx<n&&dy>=0&&dy<m&&map[dx][dy]=='W') dfs(dx,dy);
    }
    
}
int main()
{
    cin>>n>>m;
    int res=0;
    getchar();  //吸收回车符 
    for(int i=0;i<n;i++){
        for(int j=0;j<m;j++){
            scanf("%c",&map[i][j]);
        }
        getchar();// 将回车符读掉 
    }
    for(int i=0;i<n;i++)
    {
        for(int j=0;j<m;j++)
        {
            //从有水的地方开始dfs 
            if(map[i][j]=='W')
            {
                dfs(i,j);
                res++;
            }
        }
    }
    cout<<res<<endl;
    return 0;
}