题目链接传送门题意：一共有t组数据每组都n串字符串，求这n串字符串的最长公共子字符串长度，注意可以是公共子字符串是其反串。思路：数据较小，t在10以内，n在100以内，直接暴力搜索一遍ac了虽然我觉得会超时。。。注意strstr函数的最差复杂度是O(n2);代码#include#include#include#include#definemaxn100+10usingnamespacestd;ch

Substrings                        TimeLimit:1000MS    MemoryLimit:32768KB    64bitIOFormat:%I64d&%I64uDescriptionYouaregivenanumberofcase-sensitivestringsofalphabeticcharacters,findthelargeststringX,s

SubstringsTimeLimit:2000/1000MS(Java/Others)    MemoryLimit:65536/32768K(Java/Others)TotalSubmission(s):8707    AcceptedSubmission(s):4046ProblemDescriptionYouaregivenanumberofcase-sensitivestringsofa

/* * hdu1238/win.cpp * Created on: 2012-7-31 * Author : ben */ #include <cstdio> #

Substrings Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 7205 Accepted Submission(s): 3255 Problem Description You are given a number of case

http://acm.hdu.edu.cn/showproblem.php?pid=1238 Substrings Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other) Total Submission(s) : 10  &nbs

题目大意：给你n个长度小于100的字符串，让你找出他们之间的最长公共子串。其中，如果一个字符串含有另外一个子串的反串，那么这个子串也算做是他们之间的公共子串。分析：我们可以先找出给出的字符串中长度最小的一个，然后遍历它的所有子串，分别判断这个子串是不是其他字符串的子串。数据量不大，暴力可以过。实现代码如下：#include #include #include usingnamespacestd;

SubstringsTimeLimit:2000/1000MS(Java/Others)MemoryLimit:65536/32768K(Java/Others)TotalSubmission(s):7205AcceptedSubmission(s):3255ProblemDescriptionYouaregivenanumberofcase-sensitivestringsofalphabeti

SubstringsTimeLimit:2000/1000MS(Java/Others)    MemoryLimit:65536/32768K(Java/Others)TotalSubmission(s):6823    AcceptedSubmission(s):3053ProblemDescriptionYouaregivenanumberofcase-sensitivestringsofa

ProblemDescriptionYouaregivenanumberofcase-sensitivestringsofalphabeticcharacters,findthelargeststringX,suchthateitherX,oritsinversecanbefoundasasubstringofanyofthegivenstrings. InputThefirstlineofthe

http://acm.hdu.edu.cn/showproblem.php?pid=1238SubstringsTimeLimit:2000/1000ms(Java/Other)   MemoryLimit:65536/32768K(Java/Other)TotalSubmission(s):10   AcceptedSubmission(s):6Font: TimesNewRoman | Ver

                                                          SubstringsTimeLimit:2000/1000MS(Java/Others)    MemoryLimit:65536/32768K(Java/Others)TotalSubmission(s):6032    AcceptedSubmission(s):2666Prob

点击打开链接hdu1238思路：kmp+暴力枚举子串分析：1题目要求找到一个子串x，满足x或x的逆串是输入的n个字符串的子串，求最大的x，输出x的长度2题目的n最大100，每一个字符串的最大长度为100

                     这题是个搜索题。主要思想就是：首先，找到输入字符串中最短的串；然后，枚举其子串和子串的反串。这里求子串要从长度最大到最小，              因为要求的是最大子串。最后，判断枚举的子串是否是公共子串。若是，输出其长度，并结束；反之，继续求解。                     在做题过程中遇到的问题就是，把boolflag1=false,fl