Problem Description Generally speaking, there are a lot of problems about strings processing. Now you encounter another such problem. If you get two strings, such as “asdf” and “sdfg”, the result of the addition between them is “asdfg”, for “sdf” is…

题意:       给你两个字符串,让你求str1+str2,就是把1的后面和2的前面重叠的地方只显示一遍就行了 abc + bcd = abcd,要求和的长度最小,和最小的前提下求字典序最小,还有就是两个串可以交换位置的,cdab + abcd = abcdab 而不是 cdabcd,交换位置后合并是最短并且字典序最小的. 思路:       首先得到两个next数组,然后用str1 去匹配 str2,再用str2 去匹配str1,(因为可以交换),分别得到子串走到最后时匹配串的位置,x,y,…

http://acm.hdu.edu.cn/showproblem.php?pid=1403 Longest Common Substring Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 3068    Accepted Submission(s): 1087 Problem Description Given two string…

hdu 3553 Just a String (后缀数组) 题意:很简单,问一个字符串的第k大的子串是谁. 解题思路:后缀数组.先预处理一遍,把能算的都算出来.将后缀按sa排序,假如我们知道答案在那个区间范围内了(假设为[l,r]),那么我们算下这个区间内的lcp的最小值(设最小值的位置为mid,大小为x),如果x*(r-l+1)>=k,那么,答案就是这个区间的lcp的最小值的某一部分(具体是哪一部分,画个图稍微算下就出来了).如果x * ( r - l + 1 ) < k 那么我们分两种情况…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4300 Problem Description Clairewd is a member of FBI. After several years concealing in BUPT, she intercepted some important messages and she was preparing for sending it to ykwd. They had agreed that ea…

题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=1867 A + B for you again Description Generally speaking, there are a lot of problems about strings processing. Now you encounter another such problem. If you get two strings, such as “asdf” and “sdfg”, t…

题意: 给你两个字符串,输出他们合并之后的字符串,合并的时候把A的后缀和B的前缀重叠合(或者把A的前缀和B的后缀重合).要求合并后的串既包含A右包含B, 且使得合并后的字符串尽量短,其次是使得合并后的字符串字典序尽量小. 分析: 首先A和B合并他们一定是首尾重叠相连,要求合并后字典序最小,所以当合并后串长度一样时,我们要把A或B中字典序小 的放在前面. 然后计算A的后缀和B的前缀匹配长度为len1,计算A的前缀和B的后缀匹配长度为len2. 如果len1大,那么就把A放前面.如果len2大,那么…

题目链接 题目意思: 给出两个字符串a, b, 求最长的公共字串c, c是a的后缀,也是b的前缀. 本题没有具体说明哪个字符串是文本串和匹配串, 所以都要考虑 思路: 查找的时候, 当文本串结束的时候, 返回匹配串的位 /************************************************************************* > File Name: 1867.cpp > Author: Stomach_ache > Mail: sudaweit…

解题报告:给你两个字符串,让你连接起来,没有前后顺序,要求是长度最短优先,其次是字典序最小.这题我用的是KMP,做两次匹配,分别把第一次跟第二次输入的字符串放前面,然后比较两次得到的字符窜的长度和字典序. #include<cstdio> #include<cstring> +; //因为如果两个加一起就有可能超出了,干脆开两倍的数组 int next[MAX]; void get_next(const char *t) { next[] = -; int len = strlen…

A + B for you again Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4496    Accepted Submission(s): 1157 Problem Description Generally speaking, there are a lot of problems about strings process…