题意 给定字符串A.B,求其最长公共子串 后缀数组模板题,求出height数组,判断sa[i]与sa[i-1]是否分属字符串A.B,统计答案即可. #include <cstdio> #include <cstdlib> #include <cstring> #include <string> #include <algorithm> #include <iostream> using namespace std; ; char st…

Time Limit: 4000MS   Memory Limit: 131072K Total Submissions: 24756   Accepted: 10130 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: his mother is ge…

后缀数组: #include<cstdio> #include<algorithm> #include<cstring> #include<vector> using namespace std; typedef long long ll; ,mod=; char s[N]; ],Log[N]; void Sort(int* x,int* y,int m) { ; i<m; ++i)c[i]=; ; i<n; ++i)++c[x[i]]; ; i…

Long Long Message   Description The little cat is majoring in physics in the capital of Byterland. A piece of sad news comes to him these days: his mother is getting ill. Being worried about spending so much on railway tickets (Byterland is such a bi…

题目链接 题意:给两个长度不超过1e5的字符串,问两个字符串的连续公共子串最大长度为多少? 思路:两个字符串连接之后直接后缀数组+LCP,在height中找出max同时满足一左一右即可: #include<iostream> #include<cstdio> #include<cstring> #include<string.h> #include<algorithm> #include<map> #include<queue&…

题目大意:求两个字符串的最长公共子串长度 把两个串接在一起,中间放一个#,然后求出height 接下来还是老套路,二分出一个答案ans,然后去验证,如果有连续几个位置的h[i]>=ans,且存在sa[i]的最大值在第二个串里,最小值在第一个串里,说明答案成立 别再把后缀数组敲错了 #include <cmath> #include <cstdio> #include <cstring> #include <algorithm> #define ll l…

[题目分析] 用height数组RMQ的性质去求最长的公共子串. 要求sa[i]和sa[i-1]必须在两个串中,然后取height的MAX. 利用中间的字符来连接两个字符串的思想很巧妙,记得最后还需要空一个位置避免冲突. [代码] #include <cstdio> #include <cstring> #include <cmath> #include <cstdlib> #include <map> #include <set>…

BZOJ 1031 [JSOI2007]字符加密Cipher | 后缀数组模板题 将字符串复制一遍接在原串后面,然后后缀排序即可. #include <cmath> #include <cstdio> #include <cstring> #include <algorithm> #define space putchar(' ') #define enter putchar('\n') using namespace std; typedef long l…

Language: Default Long Long Message Time Limit: 4000MS   Memory Limit: 131072K Total Submissions: 21228   Accepted: 8708 Case Time Limit: 1000MS Description The little cat is majoring in physics in the capital of Byterland. A piece of sad news comes…

题意: 求两个字符串的LCP SOL: 模板题.连一起搞一搞就好了...主要是记录一下做(sha)题(bi)过程心(cao)得(dan)体(xin)会(qing) 后缀数组概念...还算是简单的,过程也非常清晰...就是老人家...马丹代码那么写意真的是...每一句代码的意思大概都知道但是不能很准确的描述...自己实现又漏洞百出...所以虽然避免了抄模板...但还是相当于一个默写的过程... 然后这个题目...非常显然嘛不是...然后就打了...然后开始调...TLE...TLE...TLE..…