字典树较简单题,无需维护标记,注意细节即可. 代码: #include <iostream> #include <cstdio> #include <cstring> using namespace std; #define N 100027 struct node { node *next[]; }*root; ][]; node *create() { node *p; p = (node *)malloc(sizeof(node)); ;i<;i++) p-…

这题我开始想的简单了,WA一次,然后看disscuss里有人说输入时长度从小到大的,然后我信了.然后开始while(1) WA;然后我尝试先放如数组.后来对了: discuss里面果然不能太相信. 根据出现的次数来判断是否为前缀. #include<stdio.h> #include<string.h> #include<stdlib.h> struct trie { trie *next[]; int sum; }; trie *root; void creattri…

题意:给你一堆字符串,问是否满足对于任意两个字符串a.b,a不是b的前缀 字典树==前缀树==Trie树 trie入门题,只用到了insert和query操作 #include <cstdio> #include <cstring> #include <iostream> using namespace std; #define maxnode 1000 #define sigma_size 30 //struct Trie //{ int ch[maxnode][si…

原题链接:http://acm.hdu.edu.cn/showproblem.php?pid=1305 字典树裸题,如下: #include<algorithm> #include<iostream> #include<cstdlib> #include<cstring> #include<cstdio> ; struct Node{ int cnt; Node *next[]; inline void set(){ cnt = ; ; i &l…

第一题:401 - Palindromes UVA : http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=96&page=show_problem&problem=342 解题思路:此题很水,只要把 mirrored string 类的对应关系搞对,基本就可以了! 但是细节要注意,首先只有一个元素的时候需要单独判断,一个字符是回文串,是不是 mirrored strin…

吐槽下我的渣渣英语啊,即使叫谷歌翻译也没有看懂,最后还是自己读了好几遍题才读懂. 题目大意:题意很简单,就是给一些互不相同的由'0','1'组成的字符串,看看有没有一个字符串是否会成为另一个的开头的子串. 直接简单粗暴的去比较就可以了. 这是原题:   Immediate Decodability  An encoding of a set of symbols is said to be immediately decodable if no code for one symbol is th…

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…

题目地址:http://poj.org/problem?id=1056 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, th…

字典树水题. #include <cstdio> #include <cstring> #include <cstdlib> typedef struct Trie { bool v; Trie *next[]; } Trie; Trie *root; bool create(char str[]) { , id; bool ret = false; Trie *p = root, *q; while (str[i]) { id = str[i] - '; ++i; i…

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…