题目链接:https://vjudge.net/problem/POJ-3450 Corporate Identity Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 8046   Accepted: 2710 Description Beside other services, ACM helps companies to clearly state their “corporate identity”, which i…

题意:多个字符串的最长公共子串. 解题关键:字符串的任何一个子串都是这个字符串的某个后缀的前缀.求A和B的最长公共子串等价于求A的后缀和B的后缀的最长公共前缀的最大值. 后缀数组的经典例题,连接在一起,二分长度,height数组遍历即可. 注意flag的问题,采用二分小于的方式,可能会出现有最优解但是flag为false的情况,下界需要-1,采用0,而采用等于的话,就不会出现,不过有些题会出现死循环. 还有因为多添加的符号一定不会加入vis数组,所以vis数组只需建立4000即可. 为什么两个字…

题目链接 题意:输入N(2 <= N <= 4000)个长度不超过200的字符串,输出字典序最小的最长公共连续子串; 思路:将所有的字符串中间加上分隔符,注:分隔符只需要和输入的字符不同,且各自不同即可,没有必要是最小的字符; 连接后缀数组求解出height之后二分长度,由于height是根据sa数组建立的,所以前面符合的就是字典序最小的,直接找到就停止即可; ps: 把之前的模板简化了下,A题才是关键; #include<iostream> #include<cstdio&…

Description Beside other services, ACM helps companies to clearly state their “corporate identity”, which includes company logo but also other signs, like trademarks. One of such companies is Internet Building Masters (IBM), which has recently asked…

Corporate Identity Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 7662   Accepted: 2644 Description Beside other services, ACM helps companies to clearly state their "corporate identity", which includes company logo but also other…

后缀数组. 解决多个字符串的最长公共子串. 采用对长度的二分,将子串按height分组,每次判断是否在每个字符串中都出现过. 复杂度O(NlogN) By:大奕哥 #include<cstring> #include<cstdio> #include<cstdlib> #include<algorithm> #include<iostream> #define rank fank using namespace std; ; int r[N],w…

传送门 求 n 个串的字典序最小的最长公共子串. 和 2 个串的处理方法差不多. 把 n 个串拼接在一起,中间连上一个没有出现过的字符防止匹配过界. 求出 height 数组后二分公共子串长度给后缀数组分组. 然后 check,每一组中是否所有的字符串都包含. 直接遍历 sa 数组,第一个满足的结果就是字典序最小的. ——代码 #include <cstdio> #include <cstring> #include <iostream> #define N 90000…

Beside other services, ACM helps companies to clearly state their “corporate identity”, which includes company logo but also other signs, like trademarks. One of such companies is Internet Building Masters (IBM), which has recently asked ACM for a he…

Corporate Identity Time Limit: 9000/3000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 3308    Accepted Submission(s): 1228 Problem Description Beside other services, ACM helps companies to clearly state their “cor…

2份模板 DC3 . 空间复杂度O3N 时间复杂度On #define F(x) ((x) / 3 + ((x) % 3 == 1 ? 0 : tb)) #define G(x) ((x) < tb ? (x) * 3 + 1 :((x) - tb) * 3 + 2) ; ; char input[MAXM]; int wa[MAXN],wb[MAXN],ws[MAXN],wv[MAXN],wsd[MAXN],r[MAXN],sa[MAXN]; char str[MAXN]; ]; int c0…