Simpsons’ Hidden Talents Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2456    Accepted Submission(s): 929 Problem Description Homer: Marge, I just figured out a way to discover some of the ta…

题意:给你两个串,问你第二个串是从第一个串的什么位置開始全然匹配的? kmp裸题,复杂度O(n+m). 当一个字符串以0为起始下标时.next[i]能够描写叙述为"不为自身的最大首尾反复子串长度". 当发生失配的情况下,j的新值next[j]取决于模式串中T[0 ~ j-1]中前缀和后缀相等部分的长度, 而且next[j]恰好等于这个最大长度. 防止超时.注意一些细节.. 另外:尽量少用strlen.变量记录下来使用比較好,用字符数组而不用string //KMP算法模板题 //hdu…

题意: 求子串w在T中出现的次数. kmp算法详解:http://www.cnblogs.com/XDJjy/p/3871045.html #include <iostream> #include <cstring> #include <string> #include <cstdio> using namespace std; int lenp; int lens; void getnext(int *next, char *p) { , k = -; n…

Clairewd’s message Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2929    Accepted Submission(s): 1132 Problem Description Clairewd is a member of FBI. After several years concealing in BUPT, s…

题目链接 题意:给定两个字符串s1,s2,求最长的s1前缀s使得s为s2的最长后缀,输出该字符串和其长度. 题解:调换s1和s2的顺序,用KMP求解即可. #include <bits/stdc++.h> using namespace std; ; char s1[N],s2[N]; int len1,len2; int next_val[N]; void getnext_val() { ,j=-; next_val[i]=-; while(i<len2){ ||s2[i]==s2[j…

Best Reward Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 500    Accepted Submission(s): 201 Problem Description After an uphill battle, General Li won a great victory. Now the head of state d…

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…

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 1711 Number Sequence (字符串匹配,KMP算法) Description Given two sequences of numbers : a1, a2, ...... , aN, and b1, b2, ...... , bM (1 <= M <= 10000, 1 <= N <= 1000000). Your task is to find a number K which make aK = b1, aK+1 = b2, ...... , aK+M…

题意是在所给的两个字符串中找最长的公共前后缀,即第一个字符串前缀和第二个字符串后缀的最长相等串. 思路是将两个字符串拼接在一起,然后直接套用 kmp 算法即可. 要注意用 next 会报编译错误,改成 Next 才过……但 next 确实不是 c++ 关键字. 代码如下: #include <iostream> #include <cstring> #include <cstdio> using namespace std; +; ],b[N]; ]; void get…