这题看是否 这题能A是侥幸,解决的办法是先存一下输入的字符串,进行排序. Problem 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 b…

Immediate Decodability Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 1378    Accepted Submission(s): 706 Problem Description An encoding of a set of symbols is said to be immediately decodabl…

题目大意:输入一系列的字符串,判断这些字符串中是否存在其中的一个字符串是另外一个字符串的前缀.. 如果是,输出Set .. is not immediately decodable 否则输出Set .. is immediately decodable 说的通俗点,就是判断一个字符串是否是两外一个字符串的前缀 解题思路: 这是一道字典树的题.一开始的时候,我用c/c++来写,然后是100行写完了,就是不知道哪里错了 这时,我实在忍不住了.直接就用java来写了 代码如下:(注意以下代码在subm…

题目链接:http://poj.org/problem?id=1056 思路: 字典树的简单应用,就是判断当前所有的单词中有木有一个是另一个的前缀,直接套用模板再在Tire定义中加一个bool类型的变量用来判断当前到达的位置是否构成另一个单词的编码 代码: #include<cstdio> #include<cstdlib> #include<iostream> #include<algorithm> #include<cstring> usin…

Immediate Decodability Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 2248    Accepted Submission(s): 1168点我 Problem Description An encoding of a set of symbols is said to be immediately decoda…

Immediate Decodability 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…

巧了,昨天刚刚写了个字典树,手到擒来,233. Problem 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 n…

前几天写好了字典,又刚好重温了KMP算法,恰逢遇到朋友吐槽最近被和谐的词越来越多了,于是突发奇想,想要自己实现一下敏感词屏蔽. 基本敏感词的屏蔽说起来很简单,只要把字符串中的敏感词替换成"***"就可以了.对于子串的查找,就KMP算法就可以了.但是敏感词这么多,总不能一个一个地遍历看看里面有没有相应的词吧! 于是我想到了前几天写的字典树.如果把它改造一下,并KMP算法结合,似乎可以节约不少时间. 首先说明一下思路: 对于KMP算法,这里不过多阐述.对于敏感词库,如果把它存进字典树,并在…

Implement a trie with insert, search, and startsWith methods. Note:You may assume that all inputs are consist of lowercase letters a-z. 这道题让我们实现一个重要但又有些复杂的数据结构-字典树, 又称前缀树或单词查找树,详细介绍可以参见网友董的博客,例如,一个保存了8个键的trie结构,"A", "to", "tea&quo…

题目链接 题意: A和B轮流在建造一个字,每次添加一个字符,要求是给定的n个串的某一个的前缀,不能添加字符的人输掉游戏,输掉的人先手下一轮的游戏.问A先手,经过k轮游戏,最后胜利的人是谁. 思路: 很显然先将n个字符串插入到字典树上,因为字典树上有分叉,不能仅仅判断字符串长度奇偶性来判断.字典树看成无环的状态图,如果按照拓扑排序逆序进行排序,在判断每个节点的时候,它的后继都已经判断过了.定义两个数组can_win[u], can_lose[u],表示u节点走下去“可能赢吗?”以及u节点走下去”可…