Problem:
 Score : 400 points    Problem Statement 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.      Constraints  2 \leq n \leq 10^5  1 \leq m \leq 10^5  1 \leq s,t \leq n  s \neq t  1 \leq u_i < v_i \leq n  1 \leq a_i,b_i \leq 10^9  If i\neq j , then u_i \neq u_j or v_i \neq v_j .  Any city can be reached from any city by changing trains.  All values in input are integers.         Input Input is given from Standard Input in the following format:  n  m  s  t  u_1  v_1  a_1  b_1  :  u_m  v_m  a_m  b_m       Output Print n lines.
In the i -th line, print 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-1 years.        Sample Input 1 4 3 2 3
1 4 1 100
1 2 1 10
1 3 20 1      Sample Output 1 999999999999998
999999999999989
999999999999979
999999999999897  After 0 years, it is optimal to do the exchange in City 1 . After 1 years, it is optimal to do the exchange in City 2 . Note that City 1 can still be visited even after the exchange office is closed.
Also note that, if it was allowed to keep 1 yen when do the exchange in City 2 and change the remaining yen into snuuk, we could reach City 3 with 999999999999998 snuuk, but this is NOT allowed. After 2 years, it is optimal to do the exchange in City 3 . After 3 years, it is optimal to do the exchange in City 4 .
Note that the same train can be used multiple times.       Sample Input 2 8 12 3 8
2 8 685087149 857180777
6 7 298270585 209942236
2 4 346080035 234079976
2 5 131857300 22507157
4 8 30723332 173476334
2 6 480845267 448565596
1 4 181424400 548830121
4 5 57429995 195056405
7 8 160277628 479932440
1 6 475692952 203530153
3 5 336869679 160714712
2 7 389775999 199123879      Sample Output 2 999999574976994
999999574976994
999999574976994
999999574976994
999999574976994
999999574976994
999999574976994
999999574976994

Solution (Python):


import sys,collections,itertools,re,math,fractions,decimal,random,array,bisect,heapq





def dijkstra(g, st):
    """Dijkstra method"""

    n = len(g)
    d = [float('inf')] * n
    d[st] = 0

    que = [(0, st)]     
    heapq.heapify(que)

    while len(que) > 0:
        dist, v0 = heapq.heappop(que)
        if d[v0] < dist: continue
        for v1, dist in g[v0]:
            if d[v1] > d[v0] + dist:
                d[v1] = d[v0] + dist
                heapq.heappush(que, (d[v1], v1))

    return d

def solve(f):
    n, m, s, t = f.read_int_list()

    g1 = [[] for _ in range(n)]
    g2 = [[] for _ in range(n)]

    for _ in xrange(m):
        u, v, a, b = f.read_int_list()
        g1[u-1].append((v-1, a))
        g1[v-1].append((u-1, a))
        g2[u-1].append((v-1, b))
        g2[v-1].append((u-1, b))

    c1 = dijkstra(g1, s-1)
    c2 = dijkstra(g2, t-1)

    ans = [0] * n
    ans[n-1] = 10**15-c1[n-1]-c2[n-1]
    for i in xrange(n-2, -1, -1):
        ans[i] = max(ans[i+1], 10**15-c1[i]-c2[i])

    return "\n".join(map(str, ans))


class Reader(object):
    def __init__(self, filename=None):
        self.file = open(filename) if filename is not None else None
        self.case = 1

    def __readline(self):
        return self.file.next().strip() if self.file else raw_input()

    def next_case(self):
        self.file.next()
        self.case += 1

    def read_int(self): return int(self.__readline())
    def read_float(self): return float(self.__readline())
    def read_long(self): return long(self.__readline())
    def read_decimal(self): return decimal.Decimal(self.__readline())
    def read_str(self): return self.__readline()

    def read_int_list(self): return map(int, self.__readline().split())
    def read_float_list(self): return map(float, self.__readline().split())
    def read_long_list(self): return map(long, self.__readline().split())
    def read_decimal_list(self): return map(decimal.Decimal, self.__readline().split())
    def read_str_list(self): return self.__readline().split()

if __name__ == '__main__':
    filename = sys.argv[1] if len(sys.argv) > 1 else None
    f = Reader(filename)
    if f.file:
        while True:
            print "Case 
            try:
                f.next_case()
            except StopIteration:
                break
    else:
        print solve(f)