如何修改BFS算法以在给定条件下找到2个顶点之间的路径？

Question

I just started to learn graphs and get stuck on this problem. I'm trying to find the shortest path(minimum number of edges) between two vertices of a graph with the condition that between every 2 intermediary vertices of the path there are 2 alternative paths (doesn't mather the lenght).

This is my BFS algorithm. Colors mean:

• white = didn't reached yet

• gray = a reached vertex will stay that color until all his neighbors won't be reached

• black = reached vertices

pi[] contains the parent of the curent vertex

void bfs(int s)
{
    int i;

    for (i=1; i<=v; i++)
    {
        if (i != s)
        {
            color[i] = WHITE;
            d[i] = INFTY;
            pi[i] = NIL;
        }
    }
    color[s] = GRAY;
    d[s] = 0;
    pi[s] = NIL;
    queueInit(&q);
    queuePush(&q,s);
    while (!queueEmpty(&q))
    {
        int u = queueFront(&q);
        int j;
        for (j=1; j<=adj[u][0]; j++)
        {
            int x = adj[u][j];
            if (color[x] == WHITE)
            {
                color[x] = GRAY;
                d[x] = d[u]+1;
                pi[x] = u;
                queuePush(&q,x);
            }
        }
        queuePop(&q);
        color[u] = BLACK;
    }
}

Please help me to change the algorithm to find the shortest path with the given condition or at least give me an advice !

Answer 1

If you start debugging your code, you'll see one of the problems is that nodes aren't popped before looping on their adjacent nodes , so after all, the algorithm is run again on them, and some of nodes are popped before the algorithm is run on them.

I also don't understand line for (j=1; j<=adj[u][0]; j++) , which is looping on adjacent vertices.

An implementation of BFS algorithms from cp-algorithms :

vector<vector<int>> adj;  // adjacency list representation
int n; // number of nodes
int s; // source vertex

queue<int> q;
vector<bool> used(n);
vector<int> d(n), p(n);

q.push(s);
used[s] = true;
p[s] = -1;
while (!q.empty()) {
    int v = q.front();
    q.pop(); // The node should be poped before looping on it's adjacent nodes
    for (int u : adj[v]) {
        if (!used[u]) {
            used[u] = true;
            q.push(u);
            d[u] = d[v] + 1;
            p[u] = v;
        }
    }
}

And then, let's say we want to print the shortest path:

if (!used[u]) {
    cout << "No path!";
} else {
    vector<int> path;
    for (int v = u; v != -1; v = p[v])
        path.push_back(v);
    reverse(path.begin(), path.end());
    cout << "Path: ";
    for (int v : path)
        cout << v << " ";
}