⚠️ 缩完之后合法顺序反而变少了

把上一节那张有环的图缩点：每个强连通分量捏成一个点，分量之间的边保留（去掉重边和自环）。n01 里那张 DAG 有 4 种合法顺序。缩点之后的这张图有几种？两个数一起输出：

G = [(0, 1), (0, 2), (1, 3), (2, 3), (3, 4), (4, 5), (5, 3), (4, 6)]
N = 7

def build(edges):
    g = {u: [] for u in range(N)}
    for a, b in edges:
        g[a].append(b)
    for u in g:
        g[u].sort()
    return g

def kosaraju(edges, second_reversed=True):
    g = build(edges)
    rg = build([(b, a) for a, b in edges])
    seen, order = set(), []
    for s in range(N):
        if s in seen:
            continue
        st = [(s, iter(g[s]))]
        seen.add(s)
        while st:
            u, it = st[-1]
            for v in it:
                if v not in seen:
                    seen.add(v)
                    st.append((v, iter(g[v])))
                    break
            else:
                order.append(u)
                st.pop()
    second = rg if second_reversed else g
    seen2, groups = set(), []
    for s in reversed(order):
        if s in seen2:
            continue
        st, grp = [s], []
        seen2.add(s)
        while st:
            u = st.pop()
            grp.append(u)
            for v in second[u]:
                if v not in seen2:
                    seen2.add(v)
                    st.append(v)
        groups.append(sorted(grp))
    return sorted(groups)

from itertools import permutations

gs = kosaraju(G)
comp = {}
for i, grp in enumerate(gs):
    for u in grp:
        comp[u] = i

ce = sorted({(comp[a], comp[b]) for a, b in G if comp[a] != comp[b]})
M = len(gs)
cnt = sum(1 for p in permutations(range(M))
          if all(p.index(a) < p.index(b) for a, b in ce))
print(str(4) + "/" + str(cnt))