Dijkstra shortest path implementation (Python recipe) by Shao-chuan Wang

ActiveState Code (http://code.activestate.com/recipes/577892/)

"""
   Copyright 2011 Shao-Chuan Wang <shaochuan.wang AT gmail.com>

    Permission is hereby granted, free of charge, to any person obtaining a copy
    of this software and associated documentation files (the "Software"), to deal
    in the Software without restriction, including without limitation the rights
    to use, copy, modify, merge, publish, distribute, sublicense, and/or sell
    copies of the Software, and to permit persons to whom the Software is
    furnished to do so, subject to the following conditions:

    The above copyright notice and this permission notice shall be included in
    all copies or substantial portions of the Software.

    THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND, EXPRESS OR
    IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF MERCHANTABILITY,
    FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT. IN NO EVENT SHALL THE
    AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER
    LIABILITY, WHETHER IN AN ACTION OF CONTRACT, TORT OR OTHERWISE, ARISING FROM,
    OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN
    THE SOFTWARE.
"""

import heapq

def dijkstra(adj, costs, s, t):
    ''' Return predecessors and min distance if there exists a shortest path 
        from s to t; Otherwise, return None '''
    Q = []     # priority queue of items; note item is mutable.
    d = {s: 0} # vertex -> minimal distance
    Qd = {}    # vertex -> [d[v], parent_v, v]
    p = {}     # predecessor
    visited_set = set([s])

    for v in adj.get(s, []):
        d[v] = costs[s, v]
        item = [d[v], s, v]
        heapq.heappush(Q, item)
        Qd[v] = item

    while Q:
        print Q
        cost, parent, u = heapq.heappop(Q)
        if u not in visited_set:
            print 'visit:', u
            p[u]= parent
            visited_set.add(u)
            if u == t:
                return p, d[u]
            for v in adj.get(u, []):
                if d.get(v):
                    if d[v] > costs[u, v] + d[u]:
                        d[v] =  costs[u, v] + d[u]
                        Qd[v][0] = d[v]    # decrease key
                        Qd[v][1] = u       # update predecessor
                        heapq._siftdown(Q, 0, Q.index(Qd[v]))
                else:
                    d[v] = costs[u, v] + d[u]
                    item = [d[v], u, v]
                    heapq.heappush(Q, item)
                    Qd[v] = item

    return None

def make_undirected(cost):
    ucost = {}
    for k, w in cost.iteritems():
        ucost[k] = w
        ucost[(k[1],k[0])] = w
    return ucost

if __name__=='__main__':

    # adjacent list
    adj = { 1: [2,3,6],
            2: [1,3,4],
            3: [1,2,4,6],
            4: [2,3,5,7],
            5: [4,6,7],
            6: [1,3,5,7],
            7: [4,5,6]}

    # edge costs
    cost = { (1,2):7,
            (1,3):9,
            (1,6):14,
            (2,3):10,
            (2,4):15,
            (3,4):11,
            (3,6):2,
            (4,5):6,
            (5,6):9,
            (4,7):2,
            (5,7):1,
            (6,7):12}

    cost = make_undirected(cost)

    s, t = 1, 7
    predecessors, min_cost = dijkstra(adj, cost, s, t)
    c = t
    path = [c]
    print 'min cost:', min_cost
    while predecessors.get(c):
        path.insert(0, predecessors[c])
        c = predecessors[c]

    print 'shortest path:', path