Hdoj 1305

原题链接

描述

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)

输入

Write a program that accepts as input a series of groups of records from 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

思路

字典树
我写的时候是默认字符串长度是递增的，但是提交之后想到万一字符串不递增呢，想着又要WA了，但却发现AC了！！！也许是题目哪里交代了，但英文题目真的懒得再看一遍了，就放过了。依然是数组形式的字典树，不过不需要find函数了，能在建树过程中确认是否符合要求，简单很多，就是输入格式有点神奇。

代码

#include <bits/stdc++.h>
#define maxn 1000000
using namespace std;

int trie[maxn][2];
int cnt[maxn], tot, f;

void create(char st[])
{
    int len = strlen(st);
    int u = 0;
    for(int i = 0; i < len; i++)
    {
        if(trie[u][st[i]-‘0‘] == 0) trie[u][st[i]-‘0‘] = ++tot;
        u = trie[u][st[i]-‘0‘];
        if(cnt[u]) f = 1;
    }
    cnt[u] = 1;
}

int main()
{
    char st[12];
    int sum = 0;
    while(scanf("%s", st)!=EOF)
    {
        if(st[0] == ‘9‘)
        {
            sum ++;
            if(f == 0) printf("Set %d is immediately decodable\n", sum);
            else printf("Set %d is not immediately decodable\n", sum);
            memset(trie, 0, sizeof(trie));
            memset(cnt, 0, sizeof(cnt));
            f = 0;
            continue;
        }
        if(f == 0) create(st);
    }
    return 0;
}