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)
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)
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).
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.
01 10 0010 0000 9 01 10 010 0000 9
Set 1 is immediately decodable Set 2 is not immediately decodable
题意: 给你一个由很多01字符串组成的集合,问你集合中有没有一个字符串是其他另外一个或多个字符串的前缀.
思路: 给节点增加个ending 标记, 记录每一个串得端点。 边插入边判断就好了
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <string>
using namespace std;
const int N = 1e5;
struct Trie{
bool ending;
int child[10];
Trie(){
ending = 0;
memset(child,0,sizeof(child));
}
}trie[N];
int trieN;
void init(){
memset(trie[0].child,0,sizeof(trie[0].child));
trieN = 0;
}
bool Insert(const string& str){
bool flag = 1;
int cur = 0;
for(int i = 0;str[i];i++){
int d = str[i] - '0';
if (trie[cur].child[d] == 0){
trie[++trieN] = Trie();
trie[cur].child[d] = trieN;
flag = 0;
}
cur = trie[cur].child[d];
if (trie[cur].ending) return true;
}
trie[cur].ending = 1;
if (flag) return true;
return false;
}
int main(){
ios::sync_with_stdio(false);
int t = 0;
init();
bool flag = 0;
string str;
while(cin >> str) {
if (str[0] != '9') {
bool ret = Insert(str);
if (flag == 0) flag = ret;
} else {
if (flag) cout << "Set " << ++t << " is not immediately decodable\n";
else cout << "Set " << ++t << " is immediately decodable\n";
init();flag = 0;
}
}
return 0;
}
本文探讨了立即可解码符号集的概念,即任一符号的编码不是其他符号编码的前缀。通过使用Trie树的数据结构,文章提供了一种高效的方法来检查一组二进制编码是否满足这一条件。
9万+

被折叠的 条评论
为什么被折叠?



