1056 IMMEDIATE DECODABILITY //trie

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最新推荐文章于 2019-06-21 18:52:30 发布

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分类专栏： ACM_数据结构文章标签： tree each insert string encoding input

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本文介绍了一种判断一组二进制编码是否具备即时可解码性的算法实现。该算法通过构建字典树来逐个检查输入的编码，确保任一编码都不是另一个编码的前缀。文章提供了一个C++实现示例，适用于处理最多八个符号的编码集合。

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IMMEDIATE DECODABILITY

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 6973		Accepted: 3341

Description

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 {A, B, C, D}

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)

Input

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).

Output

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.

Sample Input

Sample Output

Set 1 is immediately decodable
Set 2 is not immediately decodable

Source

Pacific Northwest 1998

//HDU ,PKU都过了，JOJ的没过，我表示鸭梨很大 #include<iostream> #include<string> using namespace std; struct dictree { int child[2]; bool f;/*记录当前单词出现的次数*/ } tree[700010]; int root; int nownode=0; bool insert(string source) { int len,i,j; int current,newnode; len=source.size(); if(len==0) return 1; current=root;//从根节点开始寻找 for(i=0; i<len; i++)/*逐个插入字符*/ { if(tree[current].child[source[i]-'0']!=-1)/*存在*/ { current=tree[current].child[source[i]-'0']; } else { ++nownode;//看现在第几个node还未被使用 newnode=nownode; for(j=0; j<2; j++) tree[newnode].child[j]=-1;//初始化 tree[current].child[source[i]-'0']=newnode; current=newnode; tree[current].f=false; } if(tree[current].f) return 0; } tree[current].f=true; return 1; } int main() { char temp[11]; int i; root=0; nownode=0; for(i=0; i<2; i++) tree[root].child[i]=-1;//初始化 tree[root].f=false; string a; bool flag=1; int cas=1; while(cin>>a) { if(a[0]=='9') { if(flag) { printf("Set %d is immediately decodable/n",cas++); flag=1; } else { printf("Set %d is not immediately decodable/n",cas++); flag=1; } root=0; nownode=0; for(i=0; i<2; i++) tree[root].child[i]=-1;//初始化 tree[root].f=false; } if(!flag) continue; flag=insert(a); } return 0; }