算法学习笔记（六）------深度优先搜索（DFS）

深度优先搜索（Depth-First Search）是搜索的手段之一，它从某个状态（节点）开始，不断地转移状态（节点）直到无法转移，然后回退回上一状态，继续转至其他状态。

如此不断重复，直到找到最终的解。例如求解数独，根据深度优先搜索的特点，采用递归函数实现比较简单合理。

这篇博文由于水平有限且为学习笔记，简单的写一下DFS入门知识，归纳总结留待下一篇DFS相关博文。

以经典的POJ 2386为例子。就是求出图中有几个水洼（w代表积水，.代表没有积水）。

此题较为简单，我们对图整个进行遍历，一旦发现了’W‘，我们进入搜索函数，对周边的额八个方向整体进行遍历，并同步修改图，当我们遍历完整个图的时候，我们也就计算出来了块的个数。

也就是说，有几次由函数内开始的dfs就有几处水洼。

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

#include  
#include
#include
#define N 105  
using namespace std;  

int n,m;  
char map[N][N];  
int count=0;  

void dfs(int x,int y)  
{  
     map[x][y]='.';
    for(int dx= -1;dx<=1;dx++){
        for(int dy=-1;dy<=1;dy++){
            //向x方向移动dx，向y方向移动dy，移动结果为（nx，ny）
            int nx= x+dx;
            int ny= y+dy;
            //判断该点是否在园内，以及是否有积水
            if(0<=nx&&nx0<=ny&&ny<=n&&map[nx][ny]=='w') dfs(nx,ny); 
        }
    }
    return ;
}

int main()  
{  
    cin>>n>>m;  
    for(int i=1;i<=n;i++)  
    {  
        for(int j=1;j<=m;j++)  
        {  
            if(map[i][j]=='W')  
            {  
                map[i][j]='.';  
                count++;  
                dfs(i,j);  
            }  
        }  
    }  
    cout<return 0;  
}