对于这道题,将两个字符串直接合并成为一个字符串,分别记录连个字符串结束的位置. 首先,应用黑暗圣典的模板,我们可以顺利得到height,rank,sa三个数组. 之后直接扫描1-n所有的位置,选出来一个,符合“两者都在不同子串的最大长度即可”. 此时我们会发现,sa数组记录了每个子串开头的位置,可以用于判断. #include<iostream> #include<stdio.h> #include<string> #include<string.h> #i…

Milk Patterns   Description Farmer John has noticed that the quality of milk given by his cows varies from day to day. On further investigation, he discovered that although he can't predict the quality of milk from one day to the next, there are some…

Description The 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 have been tas…

最长公共子序列(不连续) 实际问题中也有比较多的应用,比如,论文查重这种,就是很实际的一个使用方面. 这个应该是最常见的一种了,不再赘述,直接按照转移方程来进行: 按最普通的方式就是,直接构造二维矩阵,两个序列分别是Ai 以及 Bj ,c[i,j]就表示的是第一个序列的从开始到第Ai个元素,以及第二个序列的从开始到第Bj个元素,这两部分序列的最长的公共子序列,如果ai==bj,则斜对角加1,否则就是前面和上面的元素中最大的那一个,就是按照这种方式,一层层的向下递推. 最长连续公共子序列 就是st…

Long Long Message Time Limit: 4000MS   Memory Limit: 131072K Total Submissions: 26601   Accepted: 10816 Case Time Limit: 1000MS Description The little cat is majoring in physics in the capital of Byterland. A piece of sad news comes to him these days…

Milk Patterns Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 7938   Accepted: 3598 Case Time Limit: 2000MS Description Farmer John has noticed that the quality of milk given by his cows varies from day to day. On further investigation,…

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,…

Long Long Message Time Limit: 4000MS   Memory Limit: 131072K Total Submissions: 31904   Accepted: 12876 Case Time Limit: 1000MS Description The little cat is majoring in physics in the capital of Byterland. A piece of sad news comes to him these days…

好气啊,今天没有看懂后缀树和后缀自动机 只能写个后缀数组发泄一下了orz #include <cstdio> #include <cstring> *; int wa[N], wb[N], wv[N], ws[N]; int rank[N]; int height[N]; char str[N]; int s[N], sa[N]; int cmp(int *r, int a, int b, int l) { return r[a]==r[b] && r[a+l]==…

题目链接 Description 给出\(n\)个序列.找出这\(n\)个序列的最长相同子串. 在这里,相同定义为:两个子串长度相同且一个串的全部元素加上一个数就会变成另一个串. 思路 参考:hzwer. 法一:kmp 在第一个串中枚举答案串的开头位置,与其余\(n-1\)个串做\(kmp\). 法二:后缀数组 将\(n\)个串拼接起来.二分答案\(len\),将\(height\)分组,\(check\)是否有一组个数\(\geq len\)且落在\(n\)个不同的串中. 注意:\(n\)个串…