k862. 輩份比較

        假設B,C是A的祖先，D,E是A的子孫,F是E的子孫，如果我們先忽略輩分這件事情，單純從A點走到F，那就是一個單純的BFS就可以解決的事情，只要在BFS的基礎上，讓每個節點記錄自己的父輩與子孫輩的成員即可。

下列附上程式碼：

class Node:
    def __init__(self):
        self.father = []
        self.child = []
        self.dis = -99999999


s = set()
n = int(input())
d = {}
for i in range(n):
    c,f = input().split()
    if c not in d:
        d[c] = Node()
    if f not in d:
        d[f] = Node()
    d[c].father.append(f)
    d[f].child.append(c)
    
start,target = input().split()

d[start].dis = 0
p = [start]
keep = True
while keep:
    for i in p:
        for j in d[i].father:
            if j not in s:
                s.add(j)
                d[j].dis = d[i].dis-1
                p.append(j)
                if j == target:
                    keep = False
                    break
        for j in d[i].child:
            if j not in s:
                s.add(j)
                d[j].dis = d[i].dis+1
                p.append(j)
                if j == target:
                    keep = False
                    break
        if keep == False: break
print(d[target].dis)