結果

問題 No.1364 [Renaming] Road to Cherry from Zelkova
ユーザー mkawa2
提出日時 2021-10-29 16:13:44
言語 PyPy3
(7.3.15)
結果
WA  
実行時間 -
コード長 2,992 bytes
コンパイル時間 181 ms
コンパイル使用メモリ 82,432 KB
実行使用メモリ 218,996 KB
最終ジャッジ日時 2024-10-07 08:35:25
合計ジャッジ時間 23,580 ms
ジャッジサーバーID
（参考情報） 		judge4 / judge1
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 3
other AC * 19 WA * 25 RE * 1
権限があれば一括ダウンロードができます

ソースコード

diff #
プレゼンテーションモードにする

import sys
# sys.setrecursionlimit(200005)
int1 = lambda x: int(x)-1
p2D = lambda x: print(*x, sep="\n")
def II(): return int(sys.stdin.readline())
def LI(): return list(map(int, sys.stdin.readline().split()))
def LI1(): return list(map(int1, sys.stdin.readline().split()))
def LLI(rows_number): return [LI() for _ in range(rows_number)]
def LLI1(rows_number): return [LI1() for _ in range(rows_number)]
def SI(): return sys.stdin.readline().rstrip()
# dij = [(0, 1), (-1, 0), (0, -1), (1, 0)]
dij = [(0, 1), (1, 0), (0, -1), (-1, 0)]
# dij = [(0, 1), (-1, 0), (0, -1), (1, 0), (1, 1), (1, -1), (-1, 1), (-1, -1)]
inf = 1 << 63
# md = 998244353
md = 10**9+7
# 各強連結成分をリストにしてトポロジカル順に返す
def SCC(to, ot):
    n = len(to)
    # トポロジカルソート
    fin = [-1]*n
    topo = []
    for u in range(n):
        if fin[u] != -1: continue
        stack = [u]
        while stack:
            u = stack[-1]
            if fin[u] == -1:
                fin[u] = 0
                for v, _, _ in to[u]:
                    if fin[v] != -1: continue
                    stack.append(v)
            else:
                stack.pop()
                if fin[u] == 0:
                    fin[u] = 1
                    topo.append(u)
    # 逆辺でdfs
    res = []
    while topo:
        u = topo.pop()
        if fin[u] != 1: continue
        fin[u] = 2
        cur = [u]
        i = 0
        while i < len(cur):
            u = cur[i]
            for v, _, _ in ot[u]:
                if fin[v] == 2: continue
                fin[v] = 2
                cur.append(v)
            i += 1
        res.append(cur)
    return res
n, m = LI()
to = [[] for _ in range(n+1)]
ot = [[] for _ in range(n+1)]
for i in range(m):
    u, v, l, a = LI()
    to[u].append((v, l, a))
    ot[v].append((u, l, a))
gg = SCC(to, ot)
gn = len(gg)
# print(gg)
utog = [-1]*(n+1)
for g, uu in enumerate(gg):
    for u in uu:
        utog[u] = g
# print(utog)
to2 = [[] for _ in range(gn)]
ot2 = [[] for _ in range(gn)]
for u in range(n+1):
    g = utog[u]
    for v, l, a in to[u]:
        h = utog[v]
        if g == h: continue
        to2[g].append((h, l, a))
        ot2[h].append((g, l, a))
# print(to2, ot2)
def reach(g, to):
    res = [False]*gn
    res[g] = True
    stack = [g]
    while stack:
        g = stack.pop()
        for h, _, _ in to[g]:
            if res[h]: continue
            res[h] = True
            stack.append(h)
    return res
r1 = reach(utog[0], to2)
r2 = reach(utog[n], ot2)
# print(r1, r2)
cnt = []
for c0, c1 in zip(r1, r2):
    cnt.append(c0+c1)
for g in range(gn):
    if cnt[g] < 2: continue
    if len(gg[g]) > 1:
        print("INF")
        exit()
ss = [0]*gn
cc = [0]*gn
g = utog[n]
cc[g] = 1
for g in range(utog[n], utog[0]-1, -1):
    if cnt[g] < 2: continue
    for h, l, a in to[g]:
        c = cc[h]*a%md
        cc[g] += c
        ss[g] += c*l%md+ss[h]
        cc[g] %= md
        ss[g] %= md
print(ss[utog[0]])
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