poj2774 后缀数组2个字符串的最长公共子串

【poj2774 后缀数组2个字符串的最长公共子串】的更多相关文章

poj2774 后缀数组2个字符串的最长公共子串

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…

poj 2774 后缀数组两个字符串的最长公共子串

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…

POJ1226 Substrings ——后缀数组 or 暴力+strstr()函数最长公共子串

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,…

luogu 2463 [SDOI2008]Sandy的卡片 kmp || 后缀数组 n个串的最长公共子串

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…

URAL 1517 Freedom of Choice (后缀数组输出两个串最长公共子串)

版权声明:本文为博主原创文章,未经博主同意不得转载. https://blog.csdn.net/whyorwhnt/article/details/34075603 题意:给出两个串的长度(一样长),输出它们的随意一个最长公共子串 #include <cstdio> #include <cstring> #define max(a,b) ((a)>(b)?(a):(b)) #define min(a,b) ((a)<(b)? (a):(b)) const int N…

POJ2774 Long Long Message —— 后缀数组两字符串的最长公共子串

题目链接:https://vjudge.net/problem/POJ-2774 Long Long Message Time Limit: 4000MS Memory Limit: 131072K Total Submissions: 33144 Accepted: 13344 Case Time Limit: 1000MS Description The little cat is majoring in physics in the capital of Byterland. A…

求两个字符串的最长公共子串——Java实现

要求:求两个字符串的最长公共子串,如“abcdefg”和“adefgwgeweg”的最长公共子串为“defg”(子串必须是连续的) public class Main03{ // 求解两个字符号的最长公共子串 public static String maxSubstring(String strOne, String strTwo){ // 参数检查 if(strOne==null || strTwo == null){ return null; } if(strOne.equals("&qu…