b g n ac router

AC ' s StringTime limit:30000/10000 MS (java/others) Memory limit:32768/32768 K (java/others)Total Submission (s): 1127 Accepted Submission (s):problem DescriptionYou are given some words {Wi}. Then our stupid AC would give you a very long string S. AC was stupid and always wants to know whether one substring from S exists in the given words {Wi}.For example, S =

The standard KMP algorithm is used for the matching of a single pattern string, that is, to seek a pattern string match in the parent string, but now there is a problem, if multiple pattern strings are given at the same time, what should we do if we need to find the number of matches of this series of pattern strings in the parent string?Based on the KMP algorithm, we can think of a naïve algorithm is to enumerate the multiple pattern strings, and then perform multiple KMP algorithm, the process

"calculate politicization" operation print in a separate line the degree of the politicization of the given text. Print nothing for the other operations.ExamplesInput3 ·AAaAb? aaab-2? aaab-3? aaab+2? AabbaaOutput6436Test Instructions: There are two different words, and n operations. First give m words, then n operation,Operation 1, delete the word I (if it has been deleted, ignore it);Action 2, add, and vice versa.Action 3, given the string s, queries how many occurrences of the currently exist

SurfacePortalThe most violent of the bodyThe most violent method: run through all the strings that ask for the KMP and then outputJust a little bit of optimization: Save all the questions, merge the template strings the same, find next and match them.But there's no difference between the two approaches, they're violent.Not so violent.We set up an AC automaton for all the strings, sort the inquiry according to $y$, then run on the

Super Portal: http://acm.hdu.edu.cn/showproblem.php?pid=2200 Main topic: Eddy is a acmer, he not only like to do ACM problem, but also for each of the ranklist in the number of AC has a certain research, he is bored often on the paper on the ranklist of the number of each person's AC topic, and then select a part of the person (or all) Comparing the number of AC

AC Automata Initial action A common example is to give n words, and then give an article containing M characters, so you can find out how many words have appeared in the article. Of course this is not the full effect of AC automata. This article is an example, give a few words, query the text in the most occurrences of the word, if not unique, output in the order of input

string and crossdress) to build AC automata. Similarly, a contains string b how many times, is a on the AC automaton each node, how many in the B end node in the Fail tree subtree.As a result, the Dfs original tree is matched on the AC automaton, Because the chain is to turn, at the same time there is a direction, this bad treatment, so a chain to split into thr

Wireless Router introduction and wired Router Access: 1: Check whether the wired router line is set correctly. The line from the modem should be connected to the WAN interface of the router, that is, the main interface. If it is connected to another 1, 2 ....... The computer is connected to the LAN interface of other n

Give m words. A new word consisting of m words (two words can overlap) has a length of n and contains no less than k basic words. How many new words are there? M is very small, and binary is used to indicate that a new word contains a basic word. Using m words to create an AC automatic machine, you can find out the mutual inclusion of all words. The suffix feature of the AC automatic machine (the mismatche

Ultraviolet A 11468 Ac automation + dp Give k mode strings, and then randomly generate a string with a length of L. The probability of each character being selected is pi. The constructed string does not contain the probability of any mode string. Analysis: it is clear that this is the matching of the parent string of a pattern string. It is clear that an AC automatic machine needs to be built first. We use

P3808 [TEMPLATE] AC automatic machine (simple edition) and p3808 Automatic MachineBackground This is a simple template question for AC automatic machines. Used to detect correctness and algorithm constants. To prevent the card OJ, only two groups of data are supported based on the guarantee. do not submit the data maliciously.Description Given n mode strings and one text string, the number of mode strings h

input file contains two positive integers, the total number of words that the user understands (OutputAn integer that represents the total number of possible articles. You only need to know the value of the result modulus 10007.Sample Input2 2ABSample Output -DP on AC automatonDP[0][I][J] Indicates the number of strings that do not meet the conditions in the J state to the first I bitDP[1][I][J] Indicates the number of strings that meet the condition

Hdu 6096 --- String (AC automatic mechanism), hdu6096 --- string Question Link Problem DescriptionBob has a dictionary with N words in it. Now there is a list of words in which the middle part of the word has continuous letters disappeared. The middle part does not include the first and last character. We only know the prefix and suffix of each word, and the number of characters missing is uncertain, it cocould be 0. But the prefix and suffix of each

Topic Connection: HDU 3695 computer Virus on Planet PandoraTopic: Given some virus strings, it is required to judge that a given string contains several virus strings, including inversion.How to solve the problem: Expand the given string, then match it once and then invert it once.#include #include #include #include #include #include using namespace STD;Const intMAXN =250000;Const intSigma_size = -;structAho_corasick {intSZ, g[maxn][sigma_size];intTAG[MAXN], FAIL[MAXN], LAST[MAXN];intvis[ -], ju

of each register reference: http://www.cnblogs.com/ant-colonies/p/6008322.htmlThe representation of the tag register in debugax=0000 bx=0000 cx=0000 dx=0000 sp=ffee bp=0000 si=0000 di=0000ds=**** es=**** ss=**** cs=**** ip=0100NV up EI PL NZ NA PO NCThe bit flag value is1The flag value is0The logo0 CF CY (CarrY) NC (not CarrY)2 PF PE (Parity even) PO (Parity Odd)4 AF AC (auxiliary Carry) NA (No auxiliary Carry)6 ZF ZR (zero) NZ (not zero)7 SF NG (neg

asks the number of times the word appears in the article."Analysis" according to the input of the string built AC automata, constantly go fail accumulation can. Pay attention to understanding test instructions, the same string appears once only once, give several comparative pits of data:45SHEHESAYSHRHERYASHERHS 5shehesayshrheryasherhs 3sheshesheshesheshe6shehehesayshrheryasherhs Answer: 3334　　The code is as follows:1#include 2#include 3#include 4#in

problem Description"This summer vacation is not AC?" ”Yes ”"Then what do you do?" ”"Watch the World Cup, you idiot!" ”"@#$%^*% ..."Indeed, the World Cup has come, the fans of the festival has come, it is estimated that many acmer will also throw away the computer, to the TV.As fans, must want to see as much as possible complete game, of course, as a new era of good youth, you will certainly see some other programs, such as news broadcast (never forget

. The length of each word was less than and bigger than 0.4.1≤hi≤100.5. All the words in the input is different.6. All the words just consist of ' a '-' Z '. Output For each test case, the output of the string to engrave on a.If there ' s more than one possible answer, first output the shortest one. If There is still multiple solutions, output the smallest in lexicographically order.The answer is an empty string. Sample Input Sample Output Hint Compared to the normal

What to modify:1. System default Administrator account password2. System default LAN interface address3. System default DHCP and reserved address4, the system default cloud AC remote address and protocol content5. System default JS Insert Address6. System default probe position7, the default top logo, the bottom of copyright content8, WiFiDog the default parameters9, automatically fill in the MAC address WiFiDog Gateway ID, you can not enter the gatew

Si. The last line of all test case consists of M integers, while the i-th number indicates the value of Si.Technical specification1. t≤152.0 3. The length of each word was less than and bigger than 0.4.1≤hi≤100.5. All the words in the input is different.6. All the words just consist of ' a '-' Z '.Outputfor Each test case, the output of the string to engrave on a.If there ' s more than one possible answer, first output the shortest one. If There is still multiple solutions, output the smallest