这个问题,需要一组字符串求最长公共子,其实灵活运用KMP高速寻求最长前缀. 请注意,意大利愿父亲:按照输出词典的顺序的规定. 另外要提醒的是:它也被用来KMP为了解决这个问题,但是很多人认为KMP使用的暴力方法,没有真正处理的细节.发挥KMP角色.而通常这些人都大喊什么暴力法能够解决本题,没错,的确暴力法是能够解决本题的,本题的数据不大,可是请不要把KMP挂上去,然后写成暴力法了.那样会误导多少后来人啊. 建议能够主要參考我的getLongestPre这个函数,看看是怎样计算最长前缀的. 怎么推…

枚举长度最短的字符串的所有子串,再与其他串匹配. #include<cstdio> #include<cstring> #include<algorithm> #include<iostream> #include<cstdlib> #include<string> #include<cmath> #include<vector> using namespace std; ; ; ); const int in…

http://blog.sina.com.cn/s/blog_74e20d8901010pwp.html我采用的是方法三. 注意:当长度相同时,取字典序最小的. #include <iostream> #include <stdio.h> #include <string.h> #include <algorithm> /* http://blog.sina.com.cn/s/blog_74e20d8901010pwp.html 我采用的是方法三. 注意:当…

[题目链接] http://poj.org/problem?id=3450 [题目大意] 求k个字符串的最长公共子串,如果有多个答案,则输出字典序最小的. [题解] 我们对第一个串的每一个后缀和其余所有串做kmp,取匹配最小值的最大值就是答案. [代码] #include <cstring> #include <cstdio> #include <algorithm> const int N=4050,M=210; using namespace std; int nx…

题意: 给定N个字符串,寻找最长的公共字串,如果长度相同,则输出字典序最小的那个. 找其中一个字符串,枚举它的所有的字串,然后,逐个kmp比较.......相当暴力,可二分优化. #include <cstdio> #include <cmath> #include <iostream> #include <cstring> #include <string> #include <algorithm> using namespace…

题目链接:http://poj.org/problem?id=3450 题目分类:后缀数组 题意:求n个串的最长公共字串(输出字串) //#include<bits/stdc++.h> #include<stdio.h> #include<math.h> #include<algorithm> #include<string.h> using namespace std; #define N 200005 int wa[N],wb[N],wsf[…

Corporate Identity Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 5493   Accepted: 2015 Description Beside other services, ACM helps companies to clearly state their "corporate identity", which includes company logo but also other…

Description Beside other services, ACM helps companies to clearly state their “corporate identity”, which includes company logo but also other signs, like trademarks. One of such companies is Internet Building Masters (IBM), which has recently asked…

Problem Description Beside other services, ACM helps companies to clearly state their “corporate identity”, which includes company logo but also other signs, like trademarks. One of such companies is Internet Building Masters (IBM), which has recentl…

题目链接:http://poj.org/problem?id=3450 题意:给定n个字符串,求n个字符串的最长公共子串,无解输出IDENTITY LOST,否则最长的公共子串.有多组解时输出字典序最小的解 思路:后缀数组的解法,我们把n个串都链接起来,中间用一些互不相同的且都没在原串中出现过的字符来分割开.然后求后缀数组.由于求的是最长公共子串,所以我们可以二分长度x,于是问题就转变成了是否有一个长度为x的子串在n个字符串中都出现过.判断的方式是:以height数组进行分组,height值不小…

题目链接 题意:输入N(2 <= N <= 4000)个长度不超过200的字符串,输出字典序最小的最长公共连续子串; 思路:将所有的字符串中间加上分隔符,注:分隔符只需要和输入的字符不同,且各自不同即可,没有必要是最小的字符; 连接后缀数组求解出height之后二分长度,由于height是根据sa数组建立的,所以前面符合的就是字典序最小的,直接找到就停止即可; ps: 把之前的模板简化了下,A题才是关键; #include<iostream> #include<cstdio&…

这是一个典型问题KMP申请书. 结果求增加两个字符串.该法的总和是相同的前缀和后缀也是字符串的字符串,您将可以合并本节. 但是,这个问题是不是问题非常明确的含义,因为不是太清楚,外观这两个字符串的顺序无关紧要,后只需要输出的最短的组合长度的结果,并后长度一样,那么就依照字典顺序,输出字典顺序在前的字符串. 思路: 1 使用kmp在s2查找s1,那么终于结束的时候next table的值就是s1前缀和s2的后缀同样的最长的长度了. 2 输入两个字符串s1和s2.那么就能够在s2中查找s1.得到长度…

鉴于两个字符串,寻找一个字符串的频率,另一个字符串出现. 原版的kmp另一个陷阱.以下凝视了,标不是踩着好,有加班一定几率,也有机会错误,根据不同的字符串可以是详细. 变化看起来像一个,kmp速度是非常快的. #include <stdio.h> #include <string.h> const int MAX_TXT = 1000001; const int MAX_WORD = 10001; int gWlen, gTlen; char word[MAX_WORD]; int…

//截取字符串 ch 的 st~en 这一段子串返回子串的首地址 //注意用完需要根据需要最后free()掉 char* substring(char* ch,int st,int en) { ; char* pch=ch; ); pch=pch+st; ;i<length;i++) subch[i]=*(pch++); subch[length]='\0'; return subch; } 字符串截取 POJ 3080 Blue Jeans 题意 : 给出 n 个包含 60 个字符的字符串,问…

题目链接:http://poj.org/problem?id=3450 Corporate Identity Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 8549   Accepted: 2856 Description Beside other services, ACM helps companies to clearly state their “corporate identity”, which includ…

Beside other services, ACM helps companies to clearly state their “corporate identity”, which includes company logo but also other signs, like trademarks. One of such companies is Internet Building Masters (IBM), which has recently asked ACM for a he…

Beside other services, ACM helps companies to clearly state their “corporate identity”, which includes company logo but also other signs, like trademarks. One of such companies is Internet Building Masters (IBM), which has recently asked ACM for a he…

Corporate Identity Time Limit: 9000/3000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 3308    Accepted Submission(s): 1228 Problem Description Beside other services, ACM helps companies to clearly state their “cor…

http://acm.hdu.edu.cn/showproblem.php?pid=2328 Corporate Identity Time Limit: 9000/3000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 698    Accepted Submission(s): 281 Problem Description Beside other services, AC…

地址:http://acm.hdu.edu.cn/showproblem.php?pid=2328 题目: Corporate Identity Time Limit: 9000/3000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 1599    Accepted Submission(s): 614 Problem Description Beside other serv…

Beside other services, ACM helps companies to clearly state their “corporate identity”, which includes company logo but also other signs, like trademarks. One of such companies is Internet Building Masters (IBM), which has recently asked ACM for a he…

多个字符串的相关问题 这类问题的一个常用做法是,先将所有的字符串连接起来, 然后求后缀数组 和 height 数组,再利用 height 数组进行求解. 这中间可能需要二分答案. POJ - 3294 题意: 给出n个串,求至少出现在n/2+1个串中的最长公共子串 题解: (摘自罗穗骞的国家集训队论文): 将 n 个字符串连起来,中间用不相同的且没有出现在字符串中的字符隔开, 求后缀数组. 然后二分答案,用和LCP将后缀分成若干组,判断每组的后缀是否出现在不小于 k 个的原串中. 这个做法的时间…

解决问题的方法=>现象-->原因-->方案-->方案的优缺点…

近日调试一Asp.net程序,出现了“访问 IIS 元数据库失败”的错误信息,最后经过搜索发现了解决问题的方法. 解决方法如下: 1.依次点击“开始”-“运行”. 2.在“运行”栏内输入 “C:\WINDOWS\Microsoft.NET\Framework\v2.0.50727\aspnet_regiis.exe -i ”(不含引号),然后点“确定”按钮. 3.出现的cmd窗口中显示“开始安装ASP.NET XXX”等内容,等待这个窗口自动关闭. 好了,到这里一般问题就解决了,如果尚未解决请参…

Corporate Identity Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 7662   Accepted: 2644 Description Beside other services, ACM helps companies to clearly state their "corporate identity", which includes company logo but also other…

题目链接:https://vjudge.net/problem/POJ-3450 Corporate Identity Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 8046   Accepted: 2710 Description Beside other services, ACM helps companies to clearly state their “corporate identity”, which i…

Selenium私房菜系列10 -- 我遇到的问题及解决问题的方法…

Selenium私房菜系列10 -- 我遇到的问题及解决问题的方法…

题目链接:  HDU http://acm.hdu.edu.cn/showproblem.php?pid=2328 POJhttp://poj.org/problem?id=3450 #include<iostream> #include<cstring> #include<string> #include<cstdio> using namespace std; const int maxn=4444; char x[222],ans[222]; char…

传送门:http://acm.hdu.edu.cn/showproblem.php?pid=2328 题意:多组输入,n==0结束.给出n个字符串,求最长公共子串,长度相等则求字典序最小. 题解:(居然没t,可能数据水了吧)这个题和 HDU - 1238 基本一样,用string比较好操作.选第一个字符串然后两层循环(相当于找到所有的子串),然后和其他几个字符串比较看是否出现过,如果所有字符串中都出现了就记录下来,找出长度最大字典序最小的子串.否则输出"IDENTITY LOST".…