01056: IMMEDIATE DECODABILITY

trie, http://cs101.openjudge.cn/practice/01056

An encoding of a set of symbols is said to be immediately decodable if no code for one symbol is the prefix of a code for another symbol. We will assume for this problem that all codes are in binary, that no two codes within a set of codes are the same, that each code has at least one bit and no more than ten bits, and that each set has at least two codes and no more than eight.

Examples: Assume an alphabet that has symbols

The following code is immediately decodable: A:01 B:10 C:0010 D:0000

but this one is not: A:01 B:10 C:010 D:0000 (Note that A is a prefix of C)

输入

Write a program that accepts as input a series of groups of records from standard input. Each record in a group contains a collection of zeroes and ones representing a binary code for a different symbol. Each group is followed by a single separator record containing a single 9; the separator records are not part of the group. Each group is independent of other groups; the codes in one group are not related to codes in any other group (that is, each group is to be processed independently).

输出

For each group, your program should determine whether the codes in that group are immediately decodable, and should print a single output line giving the group number and stating whether the group is, or is not, immediately decodable.

样例输入

01
10
0010
0000
9
01
10
010
0000
9

样例输出

Set 1 is immediately decodable
Set 2 is not immediately decodable

来源

Pacific Northwest 1998

【陈子良 25物理学院】复习了前缀树的写法，比上次写得顺手了。我是用dfs实现的。

m=0
while True:
    try:
        a=input()
    except EOFError:
        break
    m+=1
    list1=[]
    while a!='9':
        list1.append(a)
        a=input()
    dict1={}
    flag1=True
    flag2=True
    def dfs(dict1,a,i):
        global flag1,flag2
        if not flag1:
            return
        if '-1' in dict1:
            flag1=False
            return
        if i==len(a):
            dict1['-1']=-1
            return
        if a[i] not in dict1:
            dict1[a[i]]={}
            flag2=False
        dfs(dict1[a[i]],a,i+1)
    for a in list1:
        flag2=True
        dfs(dict1,a,0)
        if not flag1:
            break
    if flag1:
        print(f'Set {m} is immediately decodable')
    else:
        print(f'Set {m} is not immediately decodable')