HDU 1238 Substing】的更多相关文章

思路: 1.找出n个字符串中最短的字符串Str[N] 2.从长到短找Str[N]的子子串 subStr[N],以及subStr[N]的反转字符串strrev(subStr[N]):(从长到短是做剪枝处理) 3.用strstr()函数遍历所有的字符串,看是否含有此子子串subStr[N]或strrev(subStr[N]):只要有一个字符串不包含subStr[N]或strrev(subStr[N])就放弃这个子串,尝试下一个: 4.找到第一个满足要求的就输出strlen(subStr[N])或者s…
http://acm.hdu.edu.cn/showproblem.php?pid=1238 Substrings Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 1238 Description You are given a number of case-sensitive strings of alphabetic characte…
题目大意:给你N个串,求出来他们的最大公共子串的长度(子串反过来也算他们的子串).   分析:很久以前就做过这道题,当时是用的strstr做的,不过相同的都是枚举了子串......还是很暴力,希望下次遇到类似的题目我已经掌握高效的方法了. ============================================================================== #include<stdio.h> #include<string.h> ; ; ; c…
Problem Description You are given a number of case-sensitive strings of alphabetic characters, find the largest string X, such that either X, or its inverse can be found as a substring of any of the given strings.   Input The first line of the input…
好吧,这题直接搜索就可以了,不过要按照长度最短的来搜,很容易想得到. 记得ACM比赛上有这道题,呃..不过,直接搜..呵呵了,真不敢想. #include <iostream> #include <cstdio> #include <string.h> #include <algorithm> using namespace std; ][]; ],s2[]; int main(){ int T; scanf("%d",&T);…
Problem 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 recentl…
传送门:http://acm.hdu.edu.cn/showproblem.php?pid=2328 题意:多组输入,n==0结束.给出n个字符串,求最长公共子串,长度相等则求字典序最小. 题解:(居然没t,可能数据水了吧)这个题和 HDU - 1238 基本一样,用string比较好操作.选第一个字符串然后两层循环(相当于找到所有的子串),然后和其他几个字符串比较看是否出现过,如果所有字符串中都出现了就记录下来,找出长度最大字典序最小的子串.否则输出"IDENTITY LOST".…
KUANGBIN带你飞 全专题整理 https://www.cnblogs.com/slzk/articles/7402292.html 专题一 简单搜索 POJ 1321 棋盘问题    //2019.3.18 POJ 2251 Dungeon Master POJ 3278 Catch That Cow  //4.8 POJ 3279 Fliptile POJ 1426 Find The Multiple  //4.8 POJ 3126 Prime Path POJ 3087 Shuffle…
首先是几份模版 KMP void kmp_pre(char x[],int m,int fail[]) { int i,j; j = fail[] = -; i = ; while (i < m) { && x[i] != x[j]) j = fail[j]; fail[++i] = ++j; } } int kmp_count(char x[],int m,char y[],int n) { ,j = ; ; while (i < n) { && y[i] !…
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