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D´Esopo-Pape algorithm

Given a graph with \(n\) vertices and \(m\) edges with weights \(w_i\) and a starting vertex \(v_0\). The task is to find the shortest path from the vertex \(v_0\) to every other vertex.

The algorithm from D´Esopo-Pape will work faster than Dijkstra's algorithm and the Bellman-Ford algorithm in most cases, and will also work for negative edges. However not for negative cycles.

Description

Let the array \(d\) contain the shortest path lengths, i.e. \(d_i\) is the current length of the shortest path from the vertex \(v_0\) to the vertex \(i\). Initially this array is filled with infinity for every vertex, except \(d_{v_0} = 0\). After the algorithm finishes, this array will contain the shortest distances.

Let the array \(p\) contain the current ancestors, i.e. \(p_i\) is the direct ancestor of the vertex \(i\) on the current shortest path from \(v_0\) to \(i\). Just like the array \(d\), the array \(p\) changes gradually during the algorithm and at the end takes its final values.

Now to the algorithm. At each step three sets of vertices are maintained:

  • \(M_0\) - vertices, for which the distance has already been calculated (although it might not be the final distance)
  • \(M_1\) - vertices, for which the distance currently is calculated
  • \(M_2\) - vertices, for which the distance has not yet been calculated

The vertices in the set \(M_1\) are stored in a bidirectional queue (deque).

At each step of the algorithm we take a vertex from the set \(M_1\) (from the front of the queue). Let \(u\) be the selected vertex. We put this vertex \(u\) into the set \(M_0\). Then we iterate over all edges coming out of this vertex. Let \(v\) be the second end of the current edge, and \(w\) its weight.

  • If \(v\) belongs to \(M_2\), then \(v\) is inserted into the set \(M_1\) by inserting it at the back of the queue. \(d_v\) is set to \(d_u + w\).
  • If \(v\) belongs to \(M_1\), then we try to improve the value of \(d_v\): \(d_v = \min(d_v, d_u + w)\). Since \(v\) is already in \(M_1\), we don't need to insert it into \(M_1\) and the queue.
  • If \(v\) belongs to \(M_0\), and if \(d_v\) can be improved \(d_v > d_u + w\), then we improve \(d_v\) and insert the vertex \(v\) back to the set \(M_1\), placing it at the beginning of the queue.

And of course, with each update in the array \(d\) we also have to update the corresponding element in the array \(p\).

Implementation

We will use an array \(m\) to store in which set each vertex is currently.

struct Edge {
    int to, w;
};

int n;
vector<vector<Edge>> adj;

const int INF = 1e9;

void shortest_paths(int v0, vector<int>& d, vector<int>& p) {
    d.assign(n, INF);
    d[v0] = 0;
    vector<int> m(n, 2);
    deque<int> q;
    q.push_back(v0);
    p.assign(n, -1);

    while (!q.empty()) {
        int u = q.front();
        q.pop_front();
        m[u] = 0;
        for (Edge e : adj[u]) {
            if (d[e.to] > d[u] + e.w) {
                d[e.to] = d[u] + e.w;
                p[e.to] = u;
                if (m[e.to] == 2) {
                    m[e.to] = 1;
                    q.push_back(e.to);
                } else if (m[e.to] == 0) {
                    m[e.to] = 1;
                    q.push_front(e.to);
                }
            }
        }
    }
}

Complexity

The algorithm performs usually quite fast. In most cases even faster than Dijkstra's algorithm. However there exist cases for which the algorithm takes exponential time.