Problem:
Kenkoooo is planning a trip in Republic of Snuke. In this country, there are n cities and m trains running. The cities are numbered 1 through n, and the i-th train connects City u_i and v_i bidirectionally. Any city can be reached from any city by changing trains.  Two currencies are used in the country: yen and snuuk. Any train fare can be paid by both yen and snuuk. The fare of the i-th train is a_i yen if paid in yen, and b_i snuuk if paid in snuuk.  In a city with a money exchange office, you can change 1 yen into 1 snuuk. However, when you do a money exchange, you have to change all your yen into snuuk. That is, if Kenkoooo does a money exchange when he has X yen, he will then have X snuuk. Currently, there is a money exchange office in every city, but the office in City i will shut down in i years and can never be used in and after that year.  Kenkoooo is planning to depart City s with 10^{15} yen in his pocket and head for City t, and change his yen into snuuk in some city while traveling. It is acceptable to do the exchange in City s or City t.  Kenkoooo would like to have as much snuuk as possible when he reaches City t by making the optimal choices for the route to travel and the city to do the exchange. For each i=0,...,n-1, find the maximum amount of snuuk that Kenkoooo has when he reaches City t if he goes on a trip from City s to City t after i years. You can assume that the trip finishes within the year.

Incorrect submission (verdict: Wrong Answer):
import heapq
def dijkstra(s):
    
    d = [float("inf")] * (n+1)
    used = [True] * (n+1) 
    d[s] = 0
    used[s] = False
    edgelist = []
    for e in graph_a[s]:
        heapq.heappush(edgelist,e)
    while len(edgelist):
        minedge = heapq.heappop(edgelist)
        
        if not used[minedge[1]]:
            continue
        v = minedge[1]
        d[v] = minedge[0]
        used[v] = False
        for e in graph_a[v]:
            if used[e[1]]:
                heapq.heappush(edgelist,[e[0]+d[v],e[1]])
    return d

def dijkstra1(s):
    
    d = [float("inf")] * (n+1)
    used = [True] * (n+1) 
    d[s] = 0
    used[s] = False
    edgelist = []
    for e in graph_b[s]:
        heapq.heappush(edgelist,e)
    while len(edgelist):
        minedge = heapq.heappop(edgelist)
        
        if not used[minedge[1]]:
            continue
        v = minedge[1]
        d[v] = minedge[0]
        used[v] = False
        for e in graph_b[v]:
            if used[e[1]]:
                heapq.heappush(edgelist,[e[0]+d[v],e[1]])
    return d


n, m, s, t = map(int, input().split())

graph_a = {i: [] for i in range(1, n + 1)}
graph_b = {i: [] for i in range(1, n + 1)}
for _ in range(m):
    u, v, a, b = map(int, input().split())
    graph_a[u].append([a, v])
    graph_a[v].append([a, u])
    graph_b[u].append([b, v])
    graph_b[v].append([b, u])

yen = dijkstra(s)
snu = dijkstra1(t)

print(yen,"\n",snu)
cost = [float("INF")] * (n + 1)
for i in range(1, n + 1):
    cost[i] = (yen[i] - yen[s]) + snu[i]

for i in range(1, n + 1):
    print(1000000000000000 - min(cost))
    del cost[1]

Fixed, accepted submission:
import heapq
INF = 10**18
def dijkstra(s, edge, n):
    d = [INF] * n
    used = [True] * n #True: not used
    d[s] = 0
    used[s] = False
    edgelist = []
    for e in edge[s]:
        heapq.heappush(edgelist,e)
    while len(edgelist):
        minedge = heapq.heappop(edgelist)
        if not used[minedge[1]]:
            continue
        v = minedge[1]
        d[v] = minedge[0]
        used[v] = False
        for e in edge[v]:
            if used[e[1]]:
                heapq.heappush(edgelist,(e[0]+d[v],e[1]))
    return d
 
n, m, s, t = map(int, input().split())
s -= 1
t -= 1
graph_a = [[] for _ in range(n)]
graph_b = [[] for _ in range(n)]
for i in range(m):
    u, v, a, b = map(int, input().split())
    u, v = u-1, v-1
    graph_a[u].append((a, v))
    graph_a[v].append((a, u))
    graph_b[u].append((b, v))
    graph_b[v].append((b, u))
 
yen = dijkstra(s, graph_a, n)
snu = dijkstra(t, graph_b, n)

cost = [0] * n
for i in range(n):
    cost[i] = (yen[i] + snu[i],i)

heapq.heapify(cost)
c = -1
aaa = 10 ** 15
for i in range(n):
    if c < i:
        while cost:
            nn, t = heapq.heappop(cost)
            if t >= i:
                c = t
                break
    print(aaa - nn)