两台各查各的

（这一节多了逐跳模拟：NODES 里每个节点有接口地址和一张小路由表；next_hop 在某节点上查表，owner 找哪个节点持有某个地址，trace 从起点一跳跳走到目标，交回 (路径, 结果)。）

贯穿拓扑：

主机 192.168.10.5/24
  └─ R1  192.168.10.1 | 10.0.1.1
       └─ R2  10.0.1.2 | 10.0.2.1
            └─ R3  10.0.2.2 | 172.16.0.1
                 └─ 目标 172.16.0.9/24

对同一个目标，在 R1 和 R3 上各查一次下一跳；再对回程的目标 192.168.10.5 查一次：

def to_int(ip):
    a, b, c, d = [int(x) for x in ip.split(".")]
    return (a << 24) | (b << 16) | (c << 8) | d


def mask(prefix):
    return (0xFFFFFFFF << (32 - prefix)) & 0xFFFFFFFF


def contains(cidr, ip):
    net, p = cidr.split("/")
    return to_int(net) & mask(int(p)) == to_int(ip) & mask(int(p))


NODES = {
    "主机": (["192.168.10.5/24"], [("0.0.0.0/0", "192.168.10.1")]),
    "R1": (["192.168.10.1/24", "10.0.1.1/24"], [("172.16.0.0/16", "10.0.1.2"), ("0.0.0.0/0", "10.0.1.2")]),
    "R2": (["10.0.1.2/24", "10.0.2.1/24"], [("172.16.0.0/16", "10.0.2.2"), ("192.168.10.0/24", "10.0.1.1")]),
    "R3": (["10.0.2.2/24", "172.16.0.1/24"], [("192.168.10.0/24", "10.0.2.1")]),
    "目标": (["172.16.0.9/24"], [("0.0.0.0/0", "172.16.0.1")]),
}


def owner(nodes, ip):
    for name, (addrs, _) in nodes.items():
        if any(a.split("/")[0] == ip for a in addrs):
            return name
    return None


def next_hop(nodes, name, target):
    addrs, table = nodes[name]
    for a in addrs:
        if contains(a, target):
            return "直连"
    best = None
    for cidr, via in table:
        if contains(cidr, target):
            plen = int(cidr.split("/")[1])
            if best is None or plen > best[0]:
                best = (plen, via)
    return best[1] if best else None


def trace(nodes, start, target, ttl=16):
    path = [start]
    cur = start
    while ttl > 0:
        hop = next_hop(nodes, cur, target)
        if hop is None:
            return path, "无路可走"
        if hop == "直连":
            dst = owner(nodes, target)
            if dst is None:
                return path, "同段但无人应答"
            path.append(dst)
            return path, "到达"
        cur = owner(nodes, hop)
        if cur is None:
            return path, "下一跳不存在"
        path.append(cur)
        ttl -= 1
    return path, "TTL耗尽"

print(next_hop(NODES, "R1", "172.16.0.9") + "/" + next_hop(NODES, "R3", "172.16.0.9") + "|" + next_hop(NODES, "R1", "192.168.10.5") + "/" + next_hop(NODES, "R3", "192.168.10.5"))