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Why does this not work:

while (!pq.empty()) {
    int curr=pq.top().second; int l=pq.top().first;
    pq.pop();
    if (curr==n-1) break;
    for (auto a: adj[curr]) {
        if (dist[a.second]>l+a.first) {
            parent[a.second]=curr;
            dist[a.second]=l+a.first;
            pq.push({dist[a.second], a.second});
        }
    }
}

and this does:

while (!pq.empty()) {
    int curr=pq.top().second; int l=pq.top().first;
    pq.pop();
    if (visited[curr]) continue;
    visited[curr]=1;
    for (auto a: adj[curr]) {
        if (dist[a.second]>l+a.first) {
            parent[a.second]=curr;
            dist[a.second]=l+a.first;
            pq.push({dist[a.second], a.second});
        }
    }
}

Shouldn't the visited array not be necessary (since we only visit each node once, the first time, where the distance is the shortest, so each successive time, if (dist[a.second]>l+a.first) does not hold and it's not added again to the queue?

question from:https://stackoverflow.com/questions/65911039/visited-array-in-dijkstras

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