C ++中圖形的BFS和DFS

Question

我一直在嘗試對圖表進行BFS和DFS處理。 我已經嘗試了所有方法，但仍然無法弄清楚我的算法出了什么問題。 請幫我解決一下這個。

我將帶有的頂點作為向量發送，該向量指向圖的頂點指向bfs和bfs，以便遍歷它並尋找正確的值。 我對此並不陌生，嘗試解決它時遇到很多問題。 如果有人可以瀏覽我的代碼，看看我的算法在哪里出錯，那將是很好的。 我希望能有新的發現。

它確實顯示了輸出，但是是錯誤的！

這是我的圖表輸入值：： 5 1、5 2、5 3、1 4、1 6

這里1是5的邊緣，2是5的邊緣，3是5的邊緣，依此類推....

這是我得到的輸出：對於BFS：5、1、2、3、4、6

對於DFS：5、4、3、2、1、5

這是我的算法：

#ifndef SORT_HPP
#define SORT_HPP

#include <string>
#include <iostream>
#include <fstream>
#include <sstream>
#include <list>
#include <vector>
#include <stack>
#include <algorithm>
#include "clsVertex.hpp"
#include "clsFileGraph.hpp"
#include "clsGraph.hpp";

class bfs : public clsGraph{
    int vert;

public:
    bfs(int s, vector<clsVertex*> verticies); //prints BFS traversal from a given source

};

class dfs: public clsGraph{
    int vert;
    list<int> adj;

public:
    dfs(int s, vector<clsVertex*> verticies);   //prints DFS traversal from a given source

};

bfs::bfs(int s, vector<clsVertex*> verticies){

    bool *visited = new bool[verticies.size()]; //creates a new boolean array of the size of the graph

    for (int i = 0; i < verticies.size(); i++){ //loops till the end of the graph
        int x = verticies[i]->ID;   //gets the value of each vertex
            //cout << "The val: " << verticies[i]->ID << endl;
            visited[x] = false; //marks that vertex as unvisited i.e: visited = false
    }

    list<int> queue;    //creates a queue
    visited[s] = true;  //marks the starting point as visited
    queue.push_back(s); //adds the starting point to the queue
    cout << endl << "The breath first sort is as follows:-" << endl << endl;
    while (queue.size() != 0){  //loops until the size of the queue is 0 
        for (int i = 0; i < verticies.size(); i++){ //loops 
            int y = verticies[i]->edges.size();
            for (int j = 0; j < y; j++){
                int z = verticies[i]->edges[j]->ID;

                if (visited[z]== false){
                    visited[z] = true;
                    queue.push_back(z);
                }

            }
        }
        cout << s << ",";
        queue.pop_front();
        if (queue.size() == 0)
            goto here;
        s = queue.front();

    }
    here:
    cout << ID << " " << graphType << endl;
    for (int i = 0; i < verticies.size(); i++){
        cout << verticies[i]->ID << "->";
        for (int j = 0; j < verticies[i]->edges.size(); j++){
            cout << verticies[i]->edges[j]->ID << endl;
        }
    }

    cout << endl << endl << "Done" << endl << endl;
}




// DFS traversal of the vertices reachable from v. It uses recursive DFSUtil()
dfs::dfs(int s, vector<clsVertex*> verticies)
{
    // Mark all the vertices as not visited
    bool *visited = new bool[verticies.size()]; //creates a new boolean array of the size of the graph

    for (int i = 0; i < verticies.size(); i++){ //loops till the end of the graph
        int x = verticies[i]->ID;   //gets the value of each vertex
        //cout << "The val: " << verticies[i]->ID << endl;
        visited[x] = false; //marks that vertex as unvisited i.e: visited = false
    }
    stack <int> depth;
    visited[s] = true;
    depth.push(s);
    //cout << s << ",";
    while (depth.size() != 0){  //loops until the size of the queue is 0 
        for (int i = 0; i < verticies.size(); i++){ //loops 
            int y = verticies[i]->edges.size();
            for (int j = 0; j < y; j++){
                int z = verticies[i]->edges[j]->ID;

                if (visited[z] == false){
                    visited[z] = true;
                    depth.push(z);
                }

            }
        }
        cout << s << ",";
        depth.pop();
        if (depth.size() == 0)
            goto there;
        s = depth.top();

    }
there:
    cout << "done";

}
#endif

Answer 1

您的BFS方法無法正常工作，因為您要向隊列中添加的所有頂點均亂序。 您應該將while ( queue.size > 0 )循環的內部替換為：

s = queue.pop_front();
for ( int i = 0; i < vertices[s]->edges.size(); i++ ) {
    int tmp = vertices[s]->edges[i]->ID;
    if ( !visited[tmp] ) {
        visited[tmp] = true;
        queue.push_back(tmp);
    }
}
cout << s << " ";

目前，您添加了頂點1的鄰居，然后添加了頂點2的鄰居，甚至沒有引用您假定的起點s。 您只需要添加當前隊列頂部哪個頂點的鄰居即可。