使用 C++ 绘制数据结构

Question

我正在制作一个将打印所有路径的控制台应用程序。但是我很难考虑如何显示从源到目的地的所有路径。

这是我的代码：

#include<iostream>

using namespace std;

int arr[8][8] = {{50,30,45,120,0,7,0,0},{30,45,28,4,70,0,0,0},
{50,20,0,38,0,0,0,0},{0,4,30,0,52,0,3,0},{0,75,0,27,0,2,0,3},
{70,0,0,0,2,0,2,0},{0,80,0,73,0,2,0,0},{60,0,90,0,30,0,0,0}};
char vertex[8] = {'A','B','C','D','E','F','G','H'};

void displayPath()
{
    system("cls");
    int start, end;
    char from, to;

    cout << "From vertex: ";
    cin >> from;
    cout << "To: ";
    cin >> to;

   switch(from)
   {
        case 'a':case 'A': start = 0; break;
        case 'b':case 'B': start = 1; break;
        case 'c':case 'C': start = 2; break;
        case 'd':case 'D': start = 3; break;
        case 'e':case 'E': start = 4; break;
        case 'f':case 'F': start = 5; break;
        case 'g':case 'G': start = 6; break;
        case 'h':case 'H': start = 7; break;    

    }

    switch(to)
   {
        case 'a':case 'A': end = 0; break;
        case 'b':case 'B': end = 1; break;
        case 'c':case 'C': end = 2; break;
        case 'd':case 'D': end = 3; break;
        case 'e':case 'E': end = 4; break;
        case 'f':case 'F': end = 5; break;
        case 'g':case 'G': end = 6; break;
        case 'h':case 'H': end = 7; break;
   }

    int temp = 0;
    int current = start;

    if(arr[start][end] > 0)
    {
        cout << vertex[start] << "->" << vertex[end];
    }
    else
        cout << "No path";
}

void computeDistance()
{
    system("cls");
    int start,end;
    char from, to;

    cout << "From vertex: ";
    cin >> from;
    cout << "To: ";
    cin >> to;

    switch(from)
    {
        case 'a':case 'A': start = 0; break;
        case 'b':case 'B': start = 1; break;
        case 'c':case 'C': start = 2; break;
        case 'd':case 'D': start = 3; break;
        case 'e':case 'E': start = 4; break;
        case 'f':case 'F': start = 5; break;
        case 'g':case 'G': start = 6; break;
        case 'h':case 'H': start = 7; break;    
    }

    switch(to)
    {
        case 'a':case 'A': end = 0; break;
        case 'b':case 'B': end = 1; break;
        case 'c':case 'C': end = 2; break;
        case 'd':case 'D': end = 3; break;
        case 'e':case 'E': end = 4; break;
        case 'f':case 'F': end = 5; break;
        case 'g':case 'G': end = 6; break;
        case 'h':case 'H': end = 7; break;
    }

    if(arr[start][end] > 0)
    {
        cout << arr[start][end] << " meters" << endl;
    }
    else
        cout << "No path";
 }

int main()
{
    int choice;

    cout << "Menu\n\n[1] Display Path\n[2] Compute Distance\n\nChoice: ";
    cin >> choice;

    switch(choice)
    {
        case 1: displayPath(); break;
        case 2: computeDistance(); break;
    }

    system("pause");
    return 0;
}

此程序仅将源提供给目标，而不是所有通过的顶点。这应该是示例输出：

From: A
To: F

A -> B -> D -> F

另外，它应该遵循最短路径的概念。并将给出总距离。我希望你能帮我解决这个问题。非常感谢您。

Answer 1

图中 2 个节点之间的各种路径的数量是指数的。这意味着它们可能有很多。 我不确定您的图表是否始终由 8 个节点构建。在以下解决方案中，我假设它是。

1. 拥有如此小的数据，您可以使用非常缓慢且效率不高的解决方案。
2. 示例：让我们生成所有可能的节点排列，并检查所有从源开始的节点，并检查它们直到它们到达目的地。
3. 检查所有路径后获取最短路径非常容易 - 只需在检查时计算每条路径的总距离。

一些代码：

#include <bits/stdc++.h>
using namespace std;

int Graph[8][8];
map <int, bool> used; // to not print same path multiple times


int valid_path(int src, int dest, vector<int>& path)
{
    int pathHash=0;
    if (path[0]!=src) return -1;
    for (size_t i=1; i<path.size(); i++)
    {
        pathHash= pathHash*10+path[i]; // such solution works only because of small size of data
        if (Graph[path[i-1]][path[i]]==0) return -1; // there is no edge between these nodes
        if (path[i]==dest) break;

    }
    return pathHash;
}

void print_path (int dest, vector<int>& path)
{
    for (size_t i=0; i<path.size(); i++)
    {
        cout << path[i] << " ";
        if (path[i]==dest) break;
        cout << " --> ";
    }
}

void generate_paths(int src, int dest)
{
    vector<int> perm {0, 1, 2, 3, 4, 5, 6, 7};
    do
    {
        int pathHash= valid_path(src, dest, perm);
        if (pathHash!=-1 && used[pathHash]==false)
        {
            used[pathHash]=true;
            print_path(dest, perm);
            cout << "\n";
        }
    } while (next_permutation(perm.begin(), perm.end()));
}


int main () {
    for (size_t i=0; i<8; i++)
    {
        for (size_t j=0; j<8; j++)
        {
            cin >> Graph[i][j];
        }
    }
    generate_paths(0, 7);
}

/* example input
0 1 0 1 0 1 0 0
0 0 1 0 0 0 0 0
0 0 0 1 0 0 0 0
0 1 0 0 1 0 0 0
0 0 0 0 0 1 0 0
0 0 0 0 0 0 1 0
0 0 0 0 0 0 0 1
0 0 0 0 0 0 0 0
*/

Answer 2

首先你需要实现一些算法来找到最短路径。
基本上有几个选择：

• Dijkstra's algorithm
• A* search algorithm

我没有看到你已经实现了任何进行此类搜索的东西。