Finding the shortest path of a graph with a cycle by breadth-first search is sometimes faster than Dijkstra's algorithm, but in the worst case it is slower.

Problem setting

Use concatenated undirected graphs ($ V $ vertices, $ E $ edges, no multiple edges, no self-loops). First, create a straight road that passes through all the points once in order from the start point, as shown in the figure below, to ensure the connection (the order of the vertices along the way is random). The other edges are randomly stretched between vertices that have no edges yet. Edge weights are randomly determined from $ 1 $ to $ 10 ^ 7 $ for each edge.

$V \in \{ 10^3,~ 10^4,~ 10^5,~ 10^6,~ 10^7 \} $ $E \in \{ 10^3,~ 10^4,~ 10^5,~ 3\cdot 10^5,~ 10^6,~ 10^7 \} $ (The reason why I put $ 3 \ cdot 10 ^ 5 $ in the number of sides is because I assumed the restrictions of the competition pro)

Time was measured for $ (V, E) $ satisfying $ V-1 \ leqq E \ leqq \ frac {V (V-1)} {2} $.

code

worstcase.py

def dijkstra0(a, in_connect_list, v, money_kind): 
    out_shortest_list = [10**15 for i in range(v)]   #Distance from vertex a
    out_shortest_list[a] = 0
    searching_list = [a]
    
    while searching_list != []:
        new_search_list = []
        for i in searching_list:
            for j in in_connect_list[i]:
                c = j[0]
                p = j[money_kind] 
                if out_shortest_list[c] > p + out_shortest_list[i]:
                    out_shortest_list[c] = p + out_shortest_list[i]
                    new_search_list.append(c)
        searching_list = list(set(new_search_list))  #Change from original code
    return out_shortest_list


def dijkstra1(s, M):
    D = [10**15] * v  #Distance from vertex s
    D[s] = 0
    V = [0] * v  #D at the apex[i]1 if is determined to be the shortest distance
    Q = []  #Priority queue
    for v in range(v):
        heapq.heappush(Q, [D[v], v])

    le = len(Q)
    while le > 0:
        q = heapq.heappop(Q)
        u = q[1]
        du = q[0]
        if V[u] == 0:
            V[u] = 1
            le -= 1
            for i in range(len(M[u])):
                v = M[u][i][0]
                luv = M[u][i][1]
                if V[v] == 0:
                    alt = du + luv
                    if D[v] > alt:
                        D[v] = alt
                        heapq.heappush(Q, [alt, v])
    return D


def dijkstra2(s, M):
    D = [10**15] * v  #Distance from vertex s
    D[s] = 0
    Q = [i for i in range(v)]
    
    while len(Q) > 0:
        mi_ind = -1
        mi = 10**15
        for i in Q:
            if D[i] < mi:
                mi = D[i]
                mi_ind = i
        Q.pop(Q.index(mi_ind))
        for i in range(len(M[mi_ind])):
            v = M[mi_ind][i][0]
            luv = M[mi_ind][i][1]
            D[v] = min(D[v], D[mi_ind] + luv)
    return D


from random import randint, random, sample
import heapq
import time

V = [10**3, 10**4, 10**5, 10**6, 10**7]
E = [10**3, 10**4, 10**5, 3 * 10**5, 10**6, 10**7] 

for v in V:
    for e in E:
        if v-1 <= e <= v*(v-1)//2:  #Conditions for concatenation and no multiple edges
            T = []
            for _ in range(10):  #Take the average measured 10 times
                
                N_shuf = [0] + sample([i for i in range(1, v)], v-1)
                M = [[] for i in range(v)]

                A = [min(N_shuf[i],N_shuf[i+1]) * v
                     + max(N_shuf[i],N_shuf[i+1]) for i in range(v-1)]
                while len(A) < e:
                    e_add = e - len(A)
                    ADD = [0] * e_add
                    i = 0
                    while i < e_add:
                        v0 = randint(0, v-1)
                        v1 = randint(0, v-1)
                        if v0 != v1:
                            ADD[i] = min(v0,v1) * v + max(v0,v1)
                            i += 1
                    A += ADD
                    A = list(set(A))  #Remove multiple edges

                for ii in A:
                    i = ii // v
                    j = ii % v
                    M[i].append([j, randint(1, 10**7)])
                    M[j].append([i, randint(1, 10**7)])

                t0 = time.time()
                
                D = dijkstra0(0, M, v, 1)
                # D = dijkstra1(0, M)
                # D = dijkstra2(0, M)

                t1 = time.time() - t0
                T.append(t1)
            print(v, e, sum(T) / 10)

result

The average of 10 times, the unit was seconds, and 2 significant figures.

- dijkstra0 </ b>: A method like breadth-first search ($ O (V ^ 2) $)

from random import randint, random, sample
import heapq
import time

for v in [1000, 5000, 10000, 50000]:
        T = []
        for _ in range(10):

            N_shuf = [0] + sample([i for i in range(1, v)], v-1)
            M = [[] for i in range(v)]

            for i in range(1, v):
                M[0].append([N_shuf[i], 10**7 - i * 2])
                M[N_shuf[i]].append([0, 10**7 - i * 2])
            for i in range(1, v-1):
                M[N_shuf[i+1]].append([N_shuf[i], 1])
                M[N_shuf[i]].append([N_shuf[i+1], 1])

            t0 = time.time()
            D = dijkstra0(0, M, v, 1)
            # D = dijkstra1(0, M)
            # D = dijkstra2(0, M)
            t1 = time.time() - t0
            T.append(t1)
        print(v, sum(T) / 10)