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…

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(…

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…

<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…

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…

题目地址:POJ 2001 考察的字典树,利用的是建树时将每个点仅仅要走过就累加.最后从根节点開始遍历,当遍历到仅仅有1次走过的时候,就说明这个地方是最短的独立前缀.然后记录下长度,输出就可以. 代码例如以下: #include <iostream> #include <stdio.h> #include <string.h> #include <stdlib.h> #include <math.h> #include <ctype.h&g…

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): 1378    Accepted Submission(s): 706 Problem Description An encoding of a set of symbols is said to be immediately decodabl…

题目链接:http://poj.org/problem?id=1204 思路分析:由于题目数据较弱,使用暴力搜索:对于所有查找的单词建立一棵字典树,在图中的每个坐标,往8个方向搜索查找即可: 需要注意的是查找时不能匹配了一个单词就不在继续往该方向查找,因为在某个坐标的某个方向上可能会匹配多个单词,所以需要一直 查找直到查找到该方向上最后一个坐标: 代码如下: #include <cstdio> #include <cstring> #include <iostream>…

题目链接:http://poj.org/problem?id=2408 World-renowned Prof. A. N. Agram's current research deals with large anagram groups. He has just found a new application for his theory on the distribution of characters in English language texts. Given such a text…

题目链接: http://poj.org/problem?id=1816 http://bailian.openjudge.cn/practice/1816?lang=en_US Time Limit: 2000MS Memory Limit: 65536K Description A word is a string of lowercases. A word pattern is a string of lowercases, '?'s and '*'s. In a pattern, a '…

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…

Phone List 时间限制:1000 ms  |  内存限制:65535 KB 难度:4   描述 Given a list of phone numbers, determine if it is consistent in the sense that no number is the prefix of another. Let's say the phone catalogue listed these numbers: Emergency 911 Alice 97 625 999…

Babelfish Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 30816   Accepted: 13283 Description You have just moved from Waterloo to a big city. The people here speak an incomprehensible dialect of a foreign language. Fortunately, you have…

题目:http://poj.org/problem?id=2513 参考博客:http://blog.csdn.net/lyy289065406/article/details/6647445 http://www.cnblogs.com/LK1994/p/3263462.html #include <iostream> #include<cstdio> #include<cstring> #include<cstdlib> #include<stac…

Time Limit: 3000MS Memory Limit: 65536K Total Submissions: 29059 Accepted: 12565 Description You have just moved from Waterloo to a big city. The people here speak an incomprehensible dialect of a foreign language. Fortunately, you have a dictionary…

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…

题目链接: http://poj.org/problem?id=3630 思路分析: 求在字符串中是否存在某个字符串为另一字符串的前缀: 即对于某个字符串而言,其是否为某个字符串的前缀,或存在某个其先前的字符串为其前缀: (1)若该字符串为某个字符串前缀,则存在一条从根节点到该字符串的最后一个字符串的路径; (2)若存在某个字符串为该字符串前缀,则在该字符串的查找路径中存在一条子路径,路径的最后的结点的endOfWord 标记为true,表示存在某个字符串为其前缀: 代码: #include <…

题目链接:http://poj.org/problem?id=2001 思路分析: 在Trie结点中添加数据域childNum,表示以该字符串为前缀的字符数目: 在创建结点时,路径上的所有除叶子节点以外的结点的childNum增加1,叶子结点的childNum设置为1: 在查询某个结点的最短前缀时: (1)若查找路径上的所有结点的childNum大于1,表明该字符串的最短路径为其自身; (2)若查找路径上存在结点的childNum等于1,表明查找的字符串是唯一以该字符串为前缀的字符串; 代码如下…

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…

Colored Sticks Time Limit: 5000MS   Memory Limit: 128000K Total Submissions: 27134   Accepted: 7186 Description You are given a bunch of wooden sticks. Each endpoint of each stick is colored with some color. Is it possible to align the sticks in a st…

487-3279 Time Limit: 2000MS        Memory Limit: 65536K Total Submissions: 309257        Accepted: 55224 Description Businesses like to have memorable telephone numbers. One way to make a telephone number memorable is to have it spell a memorable wor…

背景:为了方便九宫格手机用户发短信,希望在用户按键时,根据提供的字典(给出字符串和频数),给出各个阶段最有可能要打的单词. 题意: 首先给出的是字典,每个单词有一个出现频率.然后给出的是询问,每个询问有一个数字字符串,代表在手机上按了哪些键,以1结束.问按键的过程中最可能出现的单词分别是哪些. 思路:搞了很久.......一开始总想着以字母为各结点如何建树,询问......还是太年轻了. 以手机8个键作为字典树各节点,每个结点映射3-4对应的字母.每个结点存频率最高的串,询问的时候也可以方便的直…

Description You are given a bunch of wooden sticks. Each endpoint of each stick is colored with some color. Is it possible to align the sticks in a straight line such that the colors of the endpoints that touch are of the same color? Input Input is a…

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…

#include <iostream> #include <string> #define MAXN 50 using namespace std; struct node { node * l; node * r; bool boo; node() { l = NULL; r = NULL; boo = false; } }; bool ans; int _index; node res[*MAXN+]; node * insert(node * root,string s,in…

Phone List Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 29416   Accepted: 8774 Description Given a list of phone numbers, determine if it is consistent in the sense that no number is the prefix of another. Let's say the phone catalogu…

Shortest Prefixes Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 15610   Accepted: 6734 Description A prefix of a string is a substring starting at the beginning of the given string. The prefixes of "carbon" are: "c", &qu…

T1 IMMEDIATE DECODABILITY poj 1056 题目大意: 一些数字串 求是否存在一个串是另一个串的前缀 思路: 对于所有串经过的点权+1 如果一个点的end被访问过或经过一个被标记为end的点 就存在 #include<iostream> #include<cstdio> #include<cmath> #include<cstdlib> #include<cstring> #include<algorithm>…

POJ 排序的思想就是根据选取范围的题目的totalSubmittedNumber和totalAcceptedNumber计算一个avgAcceptRate. 每一道题都有一个value,value = acceptedNumber / avgAcceptRate + submittedNumber. 这里用到avgAcceptedRate的原因是考虑到通过的数量站的权重可能比提交的数量占更大的权重,所以给acceptedNumber乘上了一个因子. 当然计算value还有别的方法,比如POJ上…