列表给出了细分错误

Question

我不明白为什么此C ++代码会产生分段错误。

我正在尝试将数据插入邻接表

请有人告诉我如何纠正它或如何以其他方式做到这一点...

#include <iostream>
#include <list>

using namespace std;

typedef long long ll;
list<int> *adj;

void addEdge(int v, int w)
{
    adj[v].push_back(w); // Add w to v’s list. 
    adj[w].push_back(v);
}

int main()
{
    int n, m, t;
    cin >> n >> m >> t;
    int u, v;

    for (int i = 0; i < m; i++) {
        cin >> u >> v;
        addEdge(u, v);
    }

}

Answer 1

  

我该怎么办？

我会使用列表列表：

#include <cstdlib>
#include <limits>
#include <map>
#include <list>
#include <iostream>

class adjacency_list_t {
private:
    using list_type = std::map<int, std::list<int>>;
    list_type list;
public:
    adjacency_list_t() = default;

    void add_edge(int v1, int v2)
    {
        list[v1].push_back(v2);
        list[v2].push_back(v1);
    }

    std::list<int> const& get_neighbours(int v) const { return list.at(v); }

    bool are_neighbours(int v1, int v2) const
    {
        auto const &v1_list = list.at(v1);
        return std::find(v1_list.cbegin(), v1_list.cend(), v2) != v1_list.end();
    }

    friend std::ostream& operator<<(std::ostream &os, adjacency_list_t const &list)
    {
        for (auto const &v : list.list) {
            os << v.first << ' ';
            for (auto const &u : v.second) {
                os << u << ' ';
            }
            os << '\n';
        }
        return os;
    }
};

int main()
{
    std::cout << "number of edges? ";
    size_t num_edges;
    if (!(std::cin >> num_edges)) {
        std::cerr << "Input error. Bye :(\n\n";
        return EXIT_FAILURE;
    }

    adjacency_list_t list;
    for (std::size_t i{}; i < num_edges; ++i) {
        std::cout << "edge #" << i + 1 << "? ";
        int v1, v2;
        std::cin >> v1 >> v2;
        list.add_edge(v1, v2);
    }

    std::cout << "\n\n";

    bool run{ true };
    do {
        int choice;
        while (std::cout << "[1] print\n[2] print neighbours of\n"
                            "[3] are neighbours?\n[4] exit\n\n",
               !(std::cin >> choice) || choice < 1 || 4 < choice)
        {
            std::cerr << "Input error :(\n\n";
            std::cin.clear();
            std::cin.ignore(std::numeric_limits<std::streamsize>::max(), '\n');
        }

        switch (choice) {
        case 1: // print
            std::cout << "full list:\n" << list;
            break;
        case 2: // print neighbours of
        {
            std::cout << "vertex? ";
            int v;
            if (std::cin >> v) {
                auto const &neighbours{ list.get_neighbours(v) };
                for (auto const &n : neighbours)
                    std::cout << n << ' ';
            }
            break;
        }
        case 3: // are neighbours?
        {
            int v1;
            int v2;
            std::cout << "vertices? ";
            if (std::cin >> v1 >> v2) {
                std::cout << std::boolalpha << list.are_neighbours(v1, v2);
            }
            break;
        }
        case 4:
            std::cout << "Bye.";
            run = false;
        }
        std::cout << "\n\n";

    } while (run);
}

样本输出：

number of edges? 7
edge #1? 6 4
edge #2? 4 3
edge #3? 4 5
edge #4? 5 2
edge #5? 3 2
edge #6? 5 1
edge #7? 2 1


[1] print
[2] print neighbours of
[3] are neighbours?
[4] exit

1
full list:
1 5 2
2 5 3 1
3 4 2
4 6 3 5
5 4 2 1
6 4


[1] print
[2] print neighbours of
[3] are neighbours?
[4] exit

2
vertex? 3
4 2

[1] print
[2] print neighbours of
[3] are neighbours?
[4] exit

2
vertex? 5
4 2 1

[1] print
[2] print neighbours of
[3] are neighbours?
[4] exit

2
vertex? 1
5 2

[1] print
[2] print neighbours of
[3] are neighbours?
[4] exit

3
vertices? 4 3
true

[1] print
[2] print neighbours of
[3] are neighbours?
[4] exit

3
vertices? 2 6
false

[1] print
[2] print neighbours of
[3] are neighbours?
[4] exit

4
Bye.

对应图：

但是很难告诉您什么是最佳的数据结构，因为我们不知道您将使用什么数据结构。