DFS与BFS

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一、思想

DFS算法思想（递归）：

　　从某一个未访问的点作为起点开始进行搜索，一直往深处搜索未访问的点，当没有未访问的点时，则返回上一个点，重复此工作，直至所有店都被访问完。

BFS算法思想：

　　首先从一个未访问的点作为起点进行搜索，访问其相邻的所有点，再访问相邻点的所有相邻点，重复此工作，直至所有点都被访问。

二、数据结构

DFS用到的数据结构：堆栈

BFS用到的数据结构：队列

三、应用问题：DFS多用于求问题是否有解，BFS多用于求问题的最优解

四、模板

①DFS基本模板（伪代码）

void DFS(状态A)
{
    if(A不合法)
        return;
    if(A为目标状态)
        输出或记录路径
    if(A不为目标状态)
        dfs(A1) 
 }

②BFS基本模板（伪代码）

q.push(head);
while(!q.empty())
{
    temp=q.front();
    q.pop();
    if(temp?为目标状态)
        输出或记录
    if(temp不合法 )
        continue;
    if(temp合法)
        q.push(temp1); 
}

五、注意：在进栈与进队列之前要先确保栈和队列是空的，否则会引起不必要的麻烦

 六、例题分析、

题目：Oil Deposits

 The GeoSurvComp geologic survey company is responsible for detecting underground oil deposits. GeoSurvComp works with one large rectangular region of land at a time, and creates a grid that divides the land into numerous square plots. It then analyzes each plot separately, using sensing equipment to determine whether or not the plot contains oil. A plot containing oil is called a pocket. If two pockets are adjacent, then they are part of the same oil deposit. Oil deposits can be quite large and may contain numerous pockets. Your job is to determine how many different oil deposits are contained in a grid. 

输入：

The input file contains one or more grids. Each grid begins with a line containing m and n, the number of rows and columns in the grid, separated by a single space. If m = 0 it signals the end of the input; otherwise 1 <= m <= 100 and 1 <= n <= 100. Following this are m lines of n characters each (not counting the end-of-line characters). Each character corresponds to one plot, and is either `*‘, representing the absence of oil, or `@‘, representing an oil pocket. 

输出：

For each grid, output the number of distinct oil deposits. Two different pockets are part of the same oil deposit if they are adjacent horizontally, vertically, or diagonally. An oil deposit will not contain more than 100 pockets.

案例输入：

1 1
*
3 5
*@*@*
**@**
*@*@*
1 8
@@****@*
5 5 
****@
*@@*@
*@**@
@@@*@
@@**@
0 0 

案例输出：

0
1
2
2

 题目大意：

#include <iostream>
#include <cstring>
using namespace std;
 
int m,n;
char G[107][107];
int dir[8][2] = {{-1,-1},{-1,0},{-1,1},{0,-1},{0,1},{1,-1},{1,0},{1,1}};
 
void dfs(int x,int y)
{
    G[x][y] = ‘*‘;
    for(int i = 0;i < 8;i++)
    {
        int dx = x+dir[i][0];
        int dy = y+dir[i][1];
        if(G[dx][dy]==‘@‘ && dx>=1 && dx<=m && dy>=1 && dy<=n)
            dfs(dx,dy);
    }
}
 
int main()
{
    while(cin>>m>>n)
    {
        if(m == 0) break;
        int ans = 0;
        for(int i = 1;i <= m;i++)
            cin>>G[i]+1;
        for(int i = 1;i <= m;i++)
            for(int j = 1;j <= n;j++)
            {
                if(G[i][j] == ‘@‘) 
　　　　　　　　　 {
                    dfs(i,j);
                    ans++;
                }
            }
        cout<<ans<<endl;
    }
    return 0;
}

DFS与BFS

标签：技术   horizon   name   标记   return   要求   present   char   col   

原文地址：https://www.cnblogs.com/Fancy-LZC/p/9419188.html