⚠️ 两条路算分量数，必须对得上

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还是那七个点，这次把 5 → 3 那条边加回来，于是图里有了环：上面用的是 Kosaraju。再用笨办法数一遍：每一对点正反各跑一次可达性，互相到得了的归成一组。两个数输出出来对账。

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← 自己写：列出互相都走得到的点对缩完之后得到的一定是什么 →

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代码编辑器

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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)

def reach(g, s):
    seen, st = set(), [s]
    while st:
        u = st.pop()
        for v in g[u]:
            if v not in seen:
                seen.add(v)
                st.append(v)
    return seen

g = build(G)
gs = kosaraju(G)
by_scc = len(gs)

# 笨办法：每一对点都正反各跑一次可达性，自己数一遍分量
RE = {u: reach(g, u) for u in range(N)}
seen, by_brute = set(), 0
for u in range(N):
    if u in seen:
        continue
    # TODO: 把和 u 互相可达的点（含 u 自己）全部记进 seen，然后 by_brute 加 1
    pass

print(str(by_scc) + "/" + str(by_brute))

本次输入：

输出：

👩‍🏫

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⚠️ 两条路算分量数，必须对得上

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