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…

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…

UVA644 Immediate Decodability Trie Trie模板题 难度几乎相等的题P2580 于是他错误的点名开始了 对于每组数据都清空树太浪费时间,所以我们只要在需要新点时预先把新点原有的数据清空即可. 剩下除了一些细节要注意,没啥要说的了 (luogu的标签系统有些迷啊) #include<iostream> #include<cstdio> #include<cstring> using namespace std; struct data{…

题目来源:http://poj.org/problem?id=1056   http://poj.org/problem?id=3630 两题非常类似,所以在这里一并做了. 1056题目大意: 如果一组编码中不存在一个编码是另一个编码的前缀的情况,我们就称这组编码是“可立即解码的”(immediately decodable).我们假定编码都是二进制的,一组编码中没有相同的码.每个编码长度都不超过10,每组编码数目都在2到8之间. 比如,一组二进制编码:{A, B, C, D},其中:A: 01…

题意:一个码如果是另一个码的前缀,则 is not immediately decodable,不可直接解码,也就是给一串二进制数字给你,你不能对其解码,因解码出来可能有多种情况. 思路:将每个码按长度从短到长排序,逐个与其后面的码比较,若前面相同位数完全一样了,那就可以输出了. #include <iostream> #include <stdio.h> #include <string> #include <map> #include <cstri…

题意翻译 本题有多组数据.每组数据给出一列以"9"结尾的仅包含'0'和'1'的字符串,如果里面有一个是另一个的子串,输出"Set &case is not immediately decodable",否则输出"Set &case is immediately decodable".换行. case从1开始计数. 感谢@Fuko_Ibuki 提供的翻译 题目描述 PDF 输入输出格式 输入格式: 输出格式: 输入输出样例 输入样例…

IMMEDIATE DECODABILITY Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 9630   Accepted: 4555 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 sy…

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…

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…

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…

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 no two codes within a set of code…

DescriptionAn 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…

题目链接:http://poj.org/problem?id=1056 思路分析:检测某字符串是否为另一字符串的前缀,数据很弱,可以使用暴力解法.这里为了练习KMP算法使用了KMP算法. 代码如下: #include <iostream> using namespace std; ; ; char A[N][Len]; int Next[N][Len]; void get_nextval( char P[], int Next[] ) { , j = -; int PLen = strlen(…

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 sa…

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…

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…

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 no two codes within a set of code…

题目链接 题目大意 看输入的每个字符串中是否有一个字符串是另一个字符串的前缀 #include<iostream> #include<cstring> #include<algorithm> #include<string.h> #include<stdio.h> using namespace std; #define maxn 10001 string b[maxn]; ; struct ac{ ]; void init(){ sum=;fa…

<题目链接> 题目大意:给你几段只包含0,1的序列,判断这几段序列中,是否存在至少一段序列是另一段序列的前缀. 解题分析: Trie树水题,只需要在每次插入字符串,并且在Trie树上创建节点的时候,判断路径上是否已经有完整的单词出现即可. 数组版: #include <cstdio> #include <cstring> #include <algorithm> using namespace std; ]; bool flag; *][],cnt,ncas…

类似phonelist  一次ac 判断失败主要有两个要点 1. 是否包含了某段的结尾结点   说明某段被此段包含 2.此段的结尾结点是否为某段的痕迹   说明此段被包含 #include<bits/stdc++.h> using namespace std; ][]; ]; ; bool insert1( char *word ) { ; ;i<strlen(word);i++) { '; ) trie[root][ ch ]=++pos; &&root!=)retur…

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…

求这些01串是否有一个是另一个的前缀.. 就是求次数就好了嘛...emm... 网上竟然都用指针写.... #include<cstdio> #include<iostream> #include<algorithm> #include<cstring> #define maxn 2000010 #define mem(a, b) memset(a, b, sizeof(a)) using namespace std; int tot, n, m, rt;…

题目链接: http://poj.org/problem?id=1056 题意: 给定编码集, 判断它是否为可解码(没有任何一个编码是其他编码的前缀). 分析: 简单题目, 遍历一遍即可, 只需判断两个编码是否互为前缀或相等即可. 代码: #include <iostream> #include <vector> #include <string> using namespace std; string line; vector<string> vs; bo…

判断一个串是否是其他的前缀 我们需要建立一颗tire树 在插入边的时候,如果遇到一个其他串的结尾,那么就说明至少有一个串,是插入串的前缀.如果在插入完后没有新增的节点,那么插入的串就是其他串的前缀 #include<cstdio> #include<algorithm> #include<iostream> #include<cstring> using namespace std; const int manx=1<<8; char data[…

POJ1056 给定若干个字符串的集合 判断每个集合中是否有某个字符串是其他某个字符串的前缀 (哈夫曼编码有这个要求) 简单的过一遍Trie就可以了 #include<iostream> #include<cstdio> #include<cstdlib> #include<cstring> #include<cmath> #include<vector> #include<queue> #include<algor…

#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> using namespace std; char code[1000][100]; bool solve(int p) { for(int i = 1; i < p; i++) { int len = min(strlen(code[i-1]), strlen(code[i])); int j; for(…

[题目链接]: https://loj.ac/problem/10052 [题意]: 就是给一些串,是否存在两个串是相同前缀的. [题解] 模板题,不想解释了. [代码]: #include<cstdio> #include<cstring> #include<algorithm> using namespace std ; typedef long long ll; const int N = 1e4; ]; ]; int n,idx; bool f,cnt[N];…