题意:找一个出现了m次的最长子串,以及这时的最右的位置. hash的话代码还是比较好写的,,但是时间比SA多很多.. #include <stdio.h> #include <algorithm> #include <string.h> using namespace std; + ; typedef long long ll; ; ; char s[N]; int m,len,pw[N]; int H[N],pos; struct node { int id,hash…

题意:给你一个串,问期中至少出现m次的最长子串及其起始位置的坐标. 思路:hash+LCP+二分答案 #include<cstdio> #include<cstring> #include<algorithm> using namespace std; + ; ; int n, m, pos; unsigned long long H[maxn], xp[maxn]; unsigned long long hash[maxn]; int rank[maxn]; int…

题目传送门 题意:训练指南P225 分析:二分寻找长度,用hash值来比较长度为L的字串是否相等. #include <bits/stdc++.h> using namespace std; typedef unsigned long long ull; const int N = 4e4 + 5; const int x = 123; ull H[N], _hash[N], xp[N]; int rk[N]; char str[N]; int m; void get_hash(char *s…

UVA 12206 - Stammering Aliens 题目链接 题意:给定一个序列,求出出现次数大于m,长度最长的子串的最大下标 思路:后缀数组.搞出height数组后,利用二分去查找就可以 这题之前还写过hash的写法也能过,只是写后缀数组的时候,犯了一个傻逼错误,把none输出成node还一直找不到...这是刷题来第二次碰到这样的逗比错误了,还是得注意. . 代码: #include <cstdio> #include <cstring> #include <alg…

Stammering Aliens Time Limit: 2000MS   Memory Limit: 65536K       Description Dr. Ellie Arroway has established contact with an extraterrestrial civilization. However, all efforts to decode their messages have failed so far because, as luck would hav…

Dr. Ellie Arroway has established contact with an extraterrestrial civilization. However, all efforts to decode their messages have failed so far because, as luck would have it, they have stumbled upon a race of stuttering aliens! Her team has found…

题目 Source http://acm.hdu.edu.cn/showproblem.php?pid=4080 Description Dr. Ellie Arroway has established contact with an extraterrestrial civilization. However, all efforts to decode their messages have failed so far because, as luck would have it, the…

体验了一把字符串Hash的做法,感觉Hash这种人品算法好神奇. 也许这道题的正解是后缀数组,但Hash做法的优势就是编码复杂度大大降低. #include <cstdio> #include <cstring> #include <algorithm> using namespace std; + ; ; int n, m, pos; unsigned long long H[maxn], xp[maxn]; unsigned long long hash[maxn]…

1. 题目描述给定一个长为$n \in [1, 4000]$的字符串,求其中长度最长的子串,并且该子串在原串中出现至少$m$次,并求最右起始位置. 2. 基本思路两种方法:二分+后缀数组,或者二分+哈希.(1) 二分+后缀数组对子串长度进行二分,若不同后缀的公共前缀超过这个值,则对计数值累加.若计数值超过m,则证明这个公共前缀是有效的,计数过程中同时维护pos(最右边界),从而更新rpos. (2)二分+哈希仍然是对长度进行二分,然后枚举其实位置计算该长度的子串的哈希值,排序后,计算,超过m表示…

基于hash的LCP算法: #include<cstdio> #include<cstring> #include<algorithm> #define maxn 40010 using namespace std; ; int n,m,pos; unsigned long long h[maxn],xp[maxn],hash[maxn]; int rank[maxn]; bool cmp(const int &a,const int &b) { ret…