看来以后用pascal的函数要小心了: 简简单单pos其实时间复杂度是二次方级的…… 今天学习的是KMP——字符匹配算法: 这两道题也都很简单,都是为这个算法练手的, 最朴素的匹配显然是穷举起始位置然后看是否匹配,复杂度O(nm)不尽人意 kmp的思想就是尽可能利用之前匹配的信息进行匹配. 具体分析我就不讲了,传送门http://www.cppblog.com/oosky/archive/2006/07/06/9486.html讲的很详细 但据说这两题暴力都可过…… ..] of string;…

题目大意 求N个字符串的最长公共字串 题解 和POJ1226做法一样...注意是字典序最小的...WA了一次 代码: #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> using namespace std; #define MAXN 65 char p[MAXN],T[MAXN][MAXN]; int f[MAXN]; void getfail(cha…

题目链接: https://vjudge.net/problem/POJ-3080 题目大意: 找最长的公共字串(长度>=3),长度相同就找字典序最小的 解题思路: 枚举第一个串的所以子串,处理出其他串的所有子串,然后set查找,更新ans #include<iostream> #include<algorithm> #include<set> #include<string> using namespace std; int T; ]; int ma…

题目链接:https://vjudge.net/problem/POJ-1226 Substrings Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 15122   Accepted: 5309 Description You are given a number of case-sensitive strings of alphabetic characters, find the largest string X,…

题意: n<=10,len<=100 思路: 只有一个字符串的时候特判一下 #include<cstdio> #include<cstring> #include<string> #include<cmath> #include<iostream> #include<algorithm> #include<map> #include<set> #include<queue> #inclu…

题目链接:https://vjudge.net/problem/POJ-3080 Blue Jeans Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 19152   Accepted: 8524 Description The Genographic Project is a research partnership between IBM and The National Geographic Society that…

Description You are given a number of case-sensitive strings of alphabetic characters, find the largest string X, such that either X, or its inverse can be found as a substring of any of the given strings. Input The first line of the input contains a…

POJ 3080,题目链接http://poj.org/problem?id=3080 题意: 就是求m个长度为60的字符串的最长连续公共子串,2<=m<=10 规定: 1.最长公共串长度小于3输出no significant commonalities 2.若出现多个等长的最长的子串,则输出字典序最小的串 思路: 1. 求公共最小连续子串,那么先把第一个串长度>=3的所有连续子串找出来,然后由短到长查看所有主串是否有该子串. 2. 如果发现一个公共子串,那么就开始找长度+1的公共子串:…

Blue Jeans DescriptionThe Genographic Project is a research partnership between IBM and The National Geographic Society that is analyzing DNA from hundreds of thousands of contributors to map how the Earth was populated. As an IBM researcher, you hav…

题目大意 给定n个字符串,字符串可逆序可顺序,求它们的最长公共子串 题解 在输入的过程中记录一下最短的那个字符串,然后枚举起点,然后进行二分求出子串末位置,然后再验证是否是公共子串,记录最长的公共子串就是最终答案~~~~时间复杂度为O(N^3*logn) 代码: #include <iostream> #include <algorithm> #include <cstdio> #include <cstring> using namespace std;…