数据结构练习程序

喜欢ヅ旅行 2022-08-18 11:36 104阅读 0赞

今天朋友托我写一份程序，大二本科生的数据结构课程实验。要是以前的话，肯定先要定义图的邻接表结构，图的输入输出操作，图的遍历，写了很多代码。但是现在也仅仅越简洁越实用越好，这可能是由于时过境迁，人的心境也就变了。

问题描述：图的路径遍历要比结点遍历具有更为广泛的应用。写一个路径遍历算法，求出从结点L到结点I中途不过结点K的所有简单路径。（图的存储结构没有要求）

![这里写图片描述][20151214160845575]

/* ==============================================
    *   Simple path
    *   Author:WinCoder@qq.com
    *    2015.12.14
    */
    #include <iostream>
    #include <queue>
    #include <stack>
    
    using namespace std;
    
    typedef struct
    {
        char data[13];              
        int  matrix[13][13];            
    }Graph;
    
    Graph g = {
        {
       'A','B','C','D','E','F','G','H','I','J','K','L','M'},
        {
            {
       0,1,1,0,0,1,0,0,0,0,0,1,0},
            {
       1,0,1,1,0,0,1,1,0,0,0,0,1},
            {
       1,1,0,0,0,0,0,0,0,0,0,0,0},
            {
       0,1,0,0,1,0,0,0,0,0,0,0,0},
            {
       0,0,0,1,0,0,0,0,0,0,0,0,0},
            {
       1,0,0,0,0,0,0,0,0,0,0,0,0},
            {
       0,1,0,0,0,0,0,1,1,0,1,0,0},
            {
       0,1,0,0,0,0,1,0,0,0,1,0,0},
            {
       0,0,0,0,0,0,1,0,0,0,0,0,0},
            {
       0,0,0,0,0,0,0,0,0,0,0,1,1},
            {
       0,0,0,0,0,0,0,1,0,1,0,0,0},
            {
       1,0,0,0,0,0,0,0,0,1,0,0,1},
            {
       0,1,0,0,0,0,0,0,0,1,0,1,0} 
        }
    };
    
    deque<int> path;
    
    //i是出发,j是目标，k是不经过节点
    void dfs(Graph g,int i,int j,int k)
    {
        if(i == k || (path.end()!=find(path.begin(),path.end(),i)))
            return;
    
        path.push_back(i);
    
        if(i == j)
        {
            cout<<"Simple path: ";
            for(int n = 0;n< path.size(); n++)
            {   
                cout<<g.data[path[n]]<<" ";
            }
            cout<<endl;
            path.pop_back();
            return;
        }
    
        for(int n = 0;n<13;n++)
        {
            if(g.matrix[i][n] == 1)
            {   
                dfs(g,n,j,k);
            }
        }
    
        path.pop_back();
    }
    
    int main()
    {
        dfs(g,11,8,10);
        getchar();
    }

运行结果是只有8条简单路径，简单路径是指没有回路的路径。

![这里写图片描述][20151214160958521]

[20151214160845575]: /images/20220731/b9322c57a7754b3fb9a0f30c36e8d41a.png
[20151214160958521]: /images/20220731/4d0adf3452c0470a9714bd8d0ab37636.png