IMMEDIATE DECODABILITY POJ

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

0110001000009011001000009

Sample Output

Set 1 is immediately decodableSet 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;}