Submission #3044015


Source Code Expand

import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time,copy,functools

sys.setrecursionlimit(10**7)
inf = 10**20
eps = 1.0 / 10**10
mod = 10**9+7
dd = [(-1,0),(0,1),(1,0),(0,-1)]
ddn = [(-1,0),(-1,1),(0,1),(1,1),(1,0),(1,-1),(0,-1),(-1,-1)]

def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def S(): return input()
def pf(s): return print(s, flush=True)


def main():
    n,m = LI()
    s,t = LI()
    e = collections.defaultdict(list)
    for _ in range(m):
        u,v,d = LI()
        e[u].append((v,d))
        e[v].append((u,d))

    def search(s,t):
        d = collections.defaultdict(lambda: [inf, 0])
        d[s] = [0, 1]
        q = []
        heapq.heappush(q, (0, s))
        v = collections.defaultdict(bool)
        while len(q):
            k, u = heapq.heappop(q)
            if v[u]:
                continue
            v[u] = True

            if u == t:
                return d

            uc = d[u][1]

            for uv, ud in e[u]:
                if v[uv]:
                    continue
                vd = k + ud
                if d[uv][0] > vd:
                    d[uv] = [vd, uc]
                    heapq.heappush(q, (vd, uv))
                elif d[uv][0] == vd:
                    d[uv][1] += uc

        return d

    d1 = search(s,t)
    rd = d1[t][0]
    kk = rd / 2.0

    def search2(s,t):
        d = collections.defaultdict(lambda: [inf, 0])
        d[s] = [0, 1]
        q = []
        heapq.heappush(q, (0, s))
        v = collections.defaultdict(bool)
        while len(q):
            k, u = heapq.heappop(q)
            if v[u]:
                continue
            v[u] = True

            if u == t or k > kk:
                return d

            uc = d[u][1]
            dt = d1[u][0]

            for uv, ud in e[u]:
                if v[uv] or dt - ud != d1[uv][0]:
                    continue
                vd = k + ud
                if d[uv][0] > vd:
                    d[uv] = [vd, uc]
                    heapq.heappush(q, (vd, uv))
                elif d[uv][0] == vd:
                    d[uv][1] += uc

        return d


    d2 = search2(t,s)

    r = d1[t][1] ** 2 % mod
    for k in list(d1.keys()):
        t,c = d1[k]
        if t > kk:
            continue
        if t == kk:
            if d2[k][0] == t:
                r -= (c**2 * d2[k][1]**2) % mod
            continue

        for uv, ud in e[k]:
            if d2[uv][0] >= kk or t + ud + d2[uv][0] != rd:
                continue
            r -= (c**2 * d2[uv][1]**2) % mod

    return r % mod




print(main())

Submission Info

Submission Time
Task E - Avoiding Collision
User iehn
Language PyPy3 (2.4.0)
Score 0
Code Size 2975 Byte
Status TLE
Exec Time 2110 ms
Memory 128868 KB

Judge Result

Set Name Sample All
Score / Max Score 0 / 0 0 / 700
Status
AC × 4
AC × 15
TLE × 7
Set Name Test Cases
Sample sample01.txt, sample02.txt, sample03.txt, sample04.txt
All 01.txt, 02.txt, 03.txt, 04.txt, 05.txt, 06.txt, 07.txt, 08.txt, 09.txt, 10.txt, 11.txt, 12.txt, 13.txt, 14.txt, 15.txt, 16.txt, 17.txt, 18.txt, sample01.txt, sample02.txt, sample03.txt, sample04.txt
Case Name Status Exec Time Memory
01.txt AC 1147 ms 121432 KB
02.txt AC 1158 ms 117848 KB
03.txt AC 1967 ms 124004 KB
04.txt AC 1992 ms 124004 KB
05.txt AC 1278 ms 121688 KB
06.txt AC 1487 ms 124900 KB
07.txt TLE 2060 ms 123364 KB
08.txt AC 905 ms 115928 KB
09.txt AC 1792 ms 128100 KB
10.txt AC 1804 ms 126168 KB
11.txt AC 1747 ms 121444 KB
12.txt AC 1654 ms 128868 KB
13.txt TLE 2110 ms 123364 KB
14.txt TLE 2110 ms 123492 KB
15.txt TLE 2110 ms 122084 KB
16.txt TLE 2110 ms 128228 KB
17.txt TLE 2085 ms 127076 KB
18.txt TLE 2110 ms 125924 KB
sample01.txt AC 284 ms 64364 KB
sample02.txt AC 282 ms 64364 KB
sample03.txt AC 284 ms 64236 KB
sample04.txt AC 284 ms 64364 KB