Simpsons' Hidden Talents Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 4597    Accepted Submission(s): 1671 Problem Description Homer: Marge, I just figured out a way to discover some of the…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2594 题目大意:给两串字符串s1,s2,,找到最长子串满足既是s1的前缀又是s2的后缀,输出子串,及相应长度. 解题思路:这题是不是跟POJ 2752很像,没错,我们只要将s1.s2合并,不断递归直到找到长度小于等于s1.s2的公共前后缀即可. 代码 #include<iostream> #include<cstdio> #include<string> #include&…

Simpsons’ Hidden Talents Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4683    Accepted Submission(s): 1702 Problem Description Homer: Marge, I just figured out a way to discover some of the t…

Simpsons’ Hidden Talents Problem Description Write a program that, when given strings s1 and s2, finds the longest prefix of s1 that is a suffix of s2. Sample Input clinton homer riemann marjorie   Sample Output 0 rie 3   思路:要求的是s1的最长前缀是s2的后缀:那么kmp中的…

Simpsons’ Hidden Talents Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4543    Accepted Submission(s): 1648 Problem Description Homer: Marge, I just figured out a way to discover some of the t…

#include<cstdio> #include<algorithm> #include<cstring> #define N 200005 using namespace std; int buf1[N],buf2[N],sa[N],rnk[N],buc[N],n,height[N],ans,belong[N]; char s[N]; void suffix_sort() { ; ;i<m;i++) buc[i]=; ;i<=n;i++) buc[x[i…

Simpsons’ Hidden Talents Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 6888    Accepted Submission(s): 2461 Problem Description Homer: Marge, I just figured out a way to discover some of the…

HDU 2594 Simpsons’ Hidden Talents(辛普森一家的潜在天赋) Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) [Description] [题目描述] Homer: Marge, I just figured out a way to discover some of the talents we weren’t aware we had. Marge:…

Problem Description Homer: Marge, I just figured out a way to discover some of the talents we weren’t aware we had.Marge: Yeah, what is it?Homer: Take me for example. I want to find out if I have a talent in politics, OK?Marge: OK.Homer: So I take so…

Simpsons’ Hidden Talents Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1746 Accepted Submission(s): 637 Problem Description Homer: Marge, I just figured out a way to discover some of the talents…

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 10513    Accepted Submission(s): 3671 Problem Description Homer: Marge, I just figured out a way to discover some of the talents we weren’t aware…

题目链接:https://vjudge.net/problem/HDU-2594 Simpsons’ Hidden Talents Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 10647    Accepted Submission(s): 3722 Problem Description Homer: Marge, I just f…

题记: 这道题不难但是很有意思,有两种解题思路,可以说一种是横向扫描,一种是纵向扫描. 横向扫描:遍历所有字符串,每次跟当前得出的最长公共前缀串进行对比,不断修正,最后得出最长公共前缀串. 纵向扫描:对所有串,从字符串第0位开始比较,全部相等则继续比较第1,2...n位,直到发生不全部相等的情况,则得出最长公共前缀串. 横向扫描算法实现: //LeetCode_Longest Common Prefix //Written by zhou //2013.11.22 class Solution…

http://acm.hdu.edu.cn/showproblem.php?pid=2594 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 9919    Accepted Submission(s): 3418 Problem Description Homer: Marge, I just figured out a way to…

题目链接:http://acm.acmcoder.com/showproblem.php?pid=2594 题意:求最长的串 同一时候是s1的前缀又是s2的后缀.输出子串和长度. 思路:kmp 代码: #include <vector> #include <string> #include <algorithm> #include <iostream> #include <stdio.h> #include <string.h> us…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2594 这题直接用KMP算法就能够做出来,只是我还尝试了用扩展的kmp,这题用扩展的KMP效率没那么高. KMP算法: #include<stdio.h> #include<iostream> #include<string.h> using namespace std; int next[50001]; char p[50000],s[50000]; void getnex…

题意:两个字符串s.t,求s和t的最长的相同的前缀和后缀 思路:先求s的next数组,再求t的next数组(即代码中ex数组,此时不是自己与自己匹配,而是与s匹配),最后看ex[len2]即可(len2为串t的长度). #include<iostream> #include<stdio.h> #include<string.h> using namespace std; #define MaxSize 50005 int _next[MaxSize],ex[MaxSiz…

Problem Description Homer: Marge, I just figured out a way to discover some of the talents we weren’t aware we had.Marge: Yeah, what is it?Homer: Take me for example. I want to find out if I have a talent in politics, OK?Marge: OK.Homer: So I take so…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2594 思路:将两个串连起来求一遍Next数组就行长度为两者之和,遍历时注意长度应该小于两个串中的最小值 #include<cstdio> #include<iostream> #include<algorithm> #include<math.h> #include<string.h> #include<vector> #include&…

http://acm.hdu.edu.cn/showproblem.php?pid=2594 Simpsons’ Hidden Talents Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4756    Accepted Submission(s): 1732 Problem Description Homer: Marge, I j…

Simpsons' Hidden Talents Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 2798    Accepted Submission(s): 1055 Problem Description Homer: Marge, I just figured out a way to discover some of the…

Simpsons’ Hidden Talents Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 15015    Accepted Submission(s): 5151 Problem Description Homer: Marge, I just figured out a way to discover some of the…

地址:http://acm.hdu.edu.cn/showproblem.php?pid=2594 题目: Simpsons’ Hidden Talents Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 8709    Accepted Submission(s): 3051 Problem Description Homer: Mar…

Simpsons’ Hidden Talents Time Limit: 1000ms Memory Limit: 32768KB This problem will be judged on HDU. Original ID: 259464-bit integer IO format: %I64d      Java class name: Main   Homer: Marge, I just figured out a way to discover some of the talents…

Simpsons’ Hidden Talents Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2875    Accepted Submission(s): 1095 Problem Description Homer: Marge, I just figured out a way to discover some of the t…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2594 思路: 其实就是求相同的最长前缀与最长后缀 KMP算法的简单应用: 假设输入的两个字符串分别是s1,s2.s1后面再上一个任意的大写字母,然后再将串s2连在s1的后面,再在后面加上一个任意的大写字母(注意应该与前面所加的字母不同),那么对当前的s1串求其失败函数f,那么f[n-1]即为最大的匹配数,是不是很简单啊!!!至于中间和后面为什么加大些字母,留给读者自己思考咯 代码: #include…

Homer: Marge, I just figured out a way to discover some of the talents we weren’t aware we had. Marge: Yeah, what is it? Homer: Take me for example. I want to find out if I have a talent in politics, OK? Marge: OK. Homer: So I take some politician’s…

题目大意: 给出两个长度小于等于25W的字符串,求它们的最长公共子串. 题目链接:http://www.spoj.com/problems/LCS/ 算法讨论: 二分+哈希, 后缀数组, 后缀自动机. 随意做.这里面只写一下我对后缀自动机做法的理解. 首先,我们假设两个串分别为A串和B串,我们先对建立出A串的后缀自动机,然后对于B串的每一位,我们进行如下的操作:首先从第1位开始,Parent树上的位置在root,那么对于每一次操作,如果当前结点的字符可以匹配当前B串中所考虑到的字符,那么自然就l…

http://www.spoj.com/problems/LCS/ 题目:求两个串的最长公共子串 参考:https://www.cnblogs.com/autoint/p/10345276.html: 分析: 给定两个字符串 S 和 T ,求出最长公共子串,公共子串定义为在 S 和 T 中 都作为子串出现过的字符串 X . 我们为字符串 S 构造后缀自动机. 我们现在处理字符串 T ,对于每一个前缀都在 S 中寻找这个前缀的最长后缀.换句话 说,对于每个字符串 T 中的位置,我们想要找到这个位置…

题意:求n个串的最长公共子串,子串出现在一个串中可以是它的反转串出现.总长<=10^4. 题解: 对于每个串,把反转串也连进去.二分长度,分组,判断每个组. #include<cstdio> #include<cstdlib> #include<cstring> #include<iostream> using namespace std; *; int n,sl,cl,c[N],rk[N],sa[N],Rs[N],wr[N],y[N],h[N],st…