如何确定图是否具有非唯一拓扑排序

Question

我创建了一个程序，在给定图表的情况下按拓扑排序组织图表。我确定了 3 个结果：

还行
现有周期
缺少信息

前两点的输出是正确的，但第三点不正确。例如对于有 4 个顶点和边的图：1->2； 3->1； 3->4； 4->2，我得到的结果是：3 1 4 2...错了！已知的不足以得出结论。感谢任何提示或帮助，提前致谢。

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

class Graph{
    int V;
    list<int> *adj;
    public:
        Graph(int V);
        void addEdge(int u, int v);
        void topologicalSort();
};

Graph::Graph(int V){
    this->V = V;
    adj = new list<int>[V];
}

void Graph::addEdge(int u, int v){
    adj[u].push_back(v);
}

void Graph::topologicalSort(){
    vector<int> in_degree(V, 0);
    for (int u=0; u<V; u++){
        list<int>::iterator itr;
        for (itr = adj[u].begin(); itr != adj[u].end(); itr++)
             in_degree[*itr]++;}
    queue<int> q;
    for (int i = 0; i < V; i++)
        if (in_degree[i] == 0)
            q.push(i);
    int cnt = 0;
    vector <int> top_order;
    while (!q.empty()){
        int u = q.front();
        q.pop();
        top_order.push_back(u);
        list<int>::iterator itr;
        for (itr = adj[u].begin(); itr != adj[u].end(); itr++)
            if (--in_degree[*itr] == 0)
                q.push(*itr);
        cnt++;}
    if (cnt != V){
        cout << "Existing cycle\n";
        return;}
    for (int i=1; i<(int)top_order.size(); i++)
        cout << top_order[i] << " ";
    cout << endl;
}

int main(){
    setbuf(stdout, NULL);
    int N, L, u, v;
    scanf("%d %d", &N, &L);
    Graph g(N+1);
    for (int i=1; i<=L; i++){
        scanf("%d %d", &u, &v);
        g.addEdge(u, v);
    }
    g.topologicalSort();
    return 0;
}

Answer 1

要检查特定图是否具有唯一的拓扑排序，显然检查 DAG 中的哈密顿路径就足够了。引用维基百科：

If a topological sort has the property that all pairs of consecutive vertices in the sorted order are connected by edges, then these edges form a directed Hamiltonian path in the DAG. If a Hamiltonian path exists, the topological sort order is unique; no other order respects the edges of the path. Conversely, if a topological sort does not form a Hamiltonian path, the DAG will have two or more valid topological orderings, for in this case it is always possible to form a second valid ordering by swapping two consecutive vertices that are not connected by an edge to each other. Therefore, it is possible to test in linear time whether a unique ordering exists.

所以你只需要为你找到的第一个排序获取 DAG，并检查它是否形成了一条访问所有顶点的路径。

Answer 2

在给定的代码中，如果您发现自己执行了两次或更多次 q.push() 作为一次 q.pop() 的结果，那么任何结果排序都不是唯一的。检查这个可能比检查汉密尔顿路径的结果 DAG 更麻烦。

这与此处评论中讨论的情况相同：Determine whether a directed graph has a unique topological ordering?

如何确定图是否具有非唯一拓扑排序

How to determine if a graph has a non-unique topological sort

c++

sorting

algorithm

topological-sort

output