One day, Nobita found that his computer is extremely slow. After several hours' work, he finally found that it was a virus that made his poor computer slow and the virus was activated by a misoperation of opening an attachment of an email. Nobita did…

题目大意:给你一些病毒的特征码,然后再给一些文本,判断每个文本有多少种病毒,不过给的字符串都是加密处理过的,给的每个字符串都有对应一个64以内的一个数(题目里面那个表就是),然后可以把这个64以内的这个数化成6位的二进制数,然后把这些二进制数每8位再化成一个字符,这就是原来的字符,比如 QA== ->编号0 1->二进制010000 000000->每8位变成一个字符 01000000 0000(后面这4个0就是那两个==,可以舍去)-> 64('@').   分析:因为是化成8位…

最近正在学AC自动机,按照惯例需要刷一套kuangbin的AC自动机专题巩固 在网上看过很多模板,感觉kuangbin大神的模板最为简洁,于是就选择了用kuangbin大神的模板. AC自动机其实就是字典树和KMP的结合,然后去思考一下KMP的原理,然后就是在字典树上实现KMP 这里最重要的思想可能就是fail的思想,就像KMP一样,匹配失败后,有一个next的数组去回溯(最长公共前缀后缀) 如何理解了KMP的话,感觉这个不会很难理解,字典树是一个非常简单的东西就不用讲了吧. HDU - 222…

题目链接:https://vjudge.net/problem/ZOJ-3430 Detect the Virus Time Limit: 2 Seconds      Memory Limit: 65536 KB One day, Nobita found that his computer is extremely slow. After several hours' work, he finally found that it was a virus that made his poor…

传送门: Detect the Virus                                                                                                                                                                                                                                     …

Detect the Virus Time Limit: 2 Seconds      Memory Limit: 65536 KB One day, Nobita found that his computer is extremely slow. After several hours' work, he finally found that it was a virus that made his poor computer slow and the virus was activated…

题目连接:zoj 3430 Detect the Virus 题目大意:给定一个编码完的串,将每个字符相应着表的数值转换成6位二进制.然后以8为一个数值,又一次形成字符 串,推断给定询问串是否含有字符集中的串. 解题思路:主要是题意,逆编码部分注意.转换完了之后,可能有字符'\0'.所以不能用字符串的形式储存.要用int型 的数组. 注意有同样串的可能. #include <cstdio> #include <cstring> #include <queue> #inc…

Detect the Virus Time Limit: 2 Seconds      Memory Limit: 65536 KB One day, Nobita found that his computer is extremely slow. After several hours' work, he finally found that it was a virus that made his poor computer slow and the virus was activated…

题目链接:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3430 题意:给你n个编码后的模式串,和m个编码后的主串,求原来主串中含有模式串的个数 思路:首先要将模式串解码成未编码前来建立ac自动机,然后解码主串扫描统计即可. code: #include <cstdio> #include <cstring> #include <queue> #include <set> using…

解码的那些事儿,不多说. 注意解码后的结果各种情况都有,用整数数组存储,char数组会超char类型的范围(这个事最蛋疼的啊)建立自动机的时候不能用0来判断结束. #include <cstdio> #include <algorithm> #include <cstring> #include <queue> #include <string> #include <iostream> using namespace std; vec…