python-PAT甲级题解-1004

A family hierarchy is usually presented by a pedigree tree. Your job is to count those family members who have no child.

Input Specification:

Each input file contains one test case. Each case starts with a line containing 0

ID K ID[1] ID[2] ... ID[K]

where ID is a two-digit number representing a given non-leaf node, K is the number of its children, followed by a sequence of two-digit ID's of its children. For the sake of simplicity, let us fix the root ID to be 01.

The input ends with N being 0. That case must NOT be processed.

Output Specification:

For each test case, you are supposed to count those family members who have no child for every seniority level starting from the root. The numbers must be printed in a line, separated by a space, and there must be no extra space at the end of each line.

The sample case represents a tree with only 2 nodes, where 01 is the root and 02 is its only child. Hence on the root 01 level, there is 0 leaf node; and on the next level, there is 1 leaf node. Then we should output 0 1 in a line.

Sample Input:

2 1
01 1 02

结尾无空行

Sample Output:

0 1

结尾无空行

Sample Solution:

n,m=map(int,input().split())
nodes={}            #使用字典存储非叶节点
for i in range(m):
    readin=input().split()
    nodes[readin[0]]=readin[2:2+int(readin[1])]
tree=[]            #使用列表存储树

def toTree(node_list:'list(str)',layer:int):
    global nodes,tree
    if layer>len(tree):
        tree.append([])
    for i in node_list:
        tree[layer-1].append(i)
        if i in nodes:
            toTree(nodes[i],layer+1)
    return

toTree(['01'],1)
sums=[]
for i in tree:
    sum=0
    for j in i:
        if j not in nodes:
            sum+=1
    sums.append(sum)
for i in range(len(sums)-1):
    print(sums[i],end=" ")
print(sums[len(sums)-1],end="")

附提交结果