⚠️ 缩点前排不全，缩点后排得全

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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）。拿同一个 kahn_len，在原图和缩点后的图上各排一次，输出各自排出来的个数。

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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)

N = 7

def kahn_len(edges, n):
    g = {u: [] for u in range(n)}
    d = {u: 0 for u in range(n)}
    for a, b in edges:
        g[a].append(b)
        d[b] += 1
    ready = sorted(u for u in range(n) if d[u] == 0)
    out = []
    while ready:
        u = ready.pop(0)
        out.append(u)
        for v in g[u]:
            d[v] -= 1
            if d[v] == 0:
                ready.append(v)
        ready.sort()
    return len(out)

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]})

# TODO: 左边在原图上排一次（七个点），右边在缩点后的图上排一次
before, after = -1, -1

print(str(before) + "/" + str(after))

本次输入：

输出：

👩‍🏫

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