构造字符串图(Levenshtein距离)

目前，在我的计算机科学课程中，我们正在讨论图以及如何使用图找到最短距离。大约一周前，我收到了一份作业，老师给了我们一个使用整数的图的代码，我们必须调整它，以便能够使用单词列表计算Levenshtein距离。我遇到的问题是，我并不真正理解图形是如何工作的，不足以操纵它。我试着在谷歌上搜索c++中的图形，但我发现没有一个类似于我给出的程序类型。

我们刚刚完成了一个关于链表的单元，我认为图的作用与此类似。我知道每个节点将指向许多其他节点，但是在有2000个单词都指向彼此的情况下，如何在不声明结构中有那么多节点的情况下跟踪每个节点的2000个指针呢?我相信(不是100%)在我给我的程序中，我的老师使用了一个整数向量的向量来跟踪，但我不知道如何实现它。

我不要求任何人完全注释每一行，因为这是一个巨大的工作量，但如果有人能大致解释一下我将如何完成我上面的要求，也许读一下代码，给我一个粗略的理解一些章节的意思(我会在一些我特别难以理解的章节上加注释)，我将非常感激。

#include <iostream>
#include <vector>
#include <algorithm> //for max<>
#include <limits>
using namespace std;
typedef vector <int> ivec;
typedef vector <ivec> imatrix; //A vector of vectors, not how this works or how to implement
typedef vector <bool> bvec;
struct graph
{
    imatrix edges; //list of attached vertices for each node
    int numVertices;
};
//I understand the ostream overloading
ostream & operator << (ostream & stream, ivec &vec)
{
    for (int i = 0; i < vec.size(); i++)
    {
        stream << vec[i] << " ";
    }
    return stream;
}
ostream & operator << (ostream & stream, graph &g)
{
    stream << endl << "numVert = " << g.numVertices << endl;
    for (int i = 0; i < g.numVertices; i++)
    {
        stream << "vertex = " << i+1 << " | edges = " << g.edges[i] << endl;
    }
    return stream;
}
const int sentinel = -1;
bvec inTree;
ivec distanceNodes;
ivec parents;
void initGraph(graph * g);
void insertEdge(graph * g, int nodeNum, int edgeNum);
void initSearch(graph * g);
void shortestPath(graph * g, int start, int end);
int main()
{
    //I understand the main, the two numbers in insertEdge are being hooked together and the two numbers in shortestPath are what we are looking to connect in the shortest way possible
    graph g;
    initGraph(&g);
    insertEdge(&g, 1, 2);
    insertEdge(&g, 1, 3);
    insertEdge(&g, 2, 1);
    insertEdge(&g, 2, 3);
    insertEdge(&g, 2, 4);
    insertEdge(&g, 3, 1);
    insertEdge(&g, 3, 2);
    insertEdge(&g, 3, 4);
    insertEdge(&g, 4, 2);
    insertEdge(&g, 4, 3);
    insertEdge(&g, 4, 5);
    insertEdge(&g, 5, 4);
    insertEdge(&g, 6, 7);
    insertEdge(&g, 7, 6);
    cout << "The graph is " << g << endl;
    shortestPath(&g, 1, 5);
    shortestPath(&g, 2, 4);
    shortestPath(&g, 5, 2);
    shortestPath(&g, 1, 7);
    return 0;
}
void initGraph(graph * g)
{
    g -> numVertices = 0; //Why set the number of vertices to 0?
}
void insertEdge(graph * g, int nodeNum, int edgeNum)
{
    int numVertices = max(nodeNum, edgeNum); //Max finds the larger of two numbers I believe? How can this be used with strings, one is not bigger than the other
    numVertices = max(1, numVertices);
    if (numVertices > g->numVertices)
    {
        for (int i = g->numVertices; i <= numVertices; i++)
        {
            ivec nodes;
            if (g->edges.size() < i)
            {
                g -> edges.push_back(nodes);
            }
        }
        g->numVertices = numVertices;
    }
    g->edges[nodeNum - 1].push_back(edgeNum);
}
void initSearch(graph * g) //I believe this function simply resets the values from a previous search
{
    if (g == NULL)
    {
        return;
    }
    inTree.clear();
    distanceNodes.clear();
    parents.clear();
    for (int i = 0; i <= g->numVertices; i++)
    {
        inTree.push_back(false);
        distanceNodes.push_back(numeric_limits <int> :: max());
        parents.push_back(sentinel);
    }
}
void shortestPath(graph * g, int start, int end)
{
    //Very confused about how this function works
    initSearch(g);
    int edge;
    int curr; //current node
    int dist;
    distanceNodes[start] = 0; 
    curr = start;
    while (! inTree[curr])
    {
        inTree[curr] = true;
        ivec edges = g->edges[curr - 1];
        for (int i = 0; i < edges.size(); i++)
        {
            edge = edges[i];
            if (distanceNodes[edge] > distanceNodes[curr] + 1)
            {
                distanceNodes[edge] = distanceNodes[curr] + 1;
                parents[edge] = curr;
            }
        }
        curr = 1;
        dist = numeric_limits <int> :: max();
        for (int i = 1; i <= g->numVertices; i++)
        {
            if ((!inTree[i]) && (dist > distanceNodes[i]))
            {
                dist = distanceNodes[i];
                curr = i;
            }
        }
    }
    ivec path;
    if (distanceNodes[end] == numeric_limits <int> :: max()) //is there a numeric_limits <string> :: max?
    {
        cout << "No way from " << start << " to " << end << endl;
    }
    else
    {
        int temp = end;
        while (temp != start)
        {
            path.push_back(temp);
            temp = parents[temp];
        }
        path.push_back(start);
        reverse(path.begin(), path.end());
        cout << "From " << start << " to " << end << " is " << path << endl;
    }
}

如果你能帮忙，那将是最受欢迎的，因为我很可能会有更多的项目与图表，我正在努力，因为不理解他们。

谢谢你,特里斯坦