Corporate Identity Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 5493   Accepted: 2015 Description Beside other services, ACM helps companies to clearly state their "corporate identity", which includes company logo but also other…

这个问题,需要一组字符串求最长公共子,其实灵活运用KMP高速寻求最长前缀. 请注意,意大利愿父亲:按照输出词典的顺序的规定. 另外要提醒的是:它也被用来KMP为了解决这个问题,但是很多人认为KMP使用的暴力方法,没有真正处理的细节.发挥KMP角色.而通常这些人都大喊什么暴力法能够解决本题,没错,的确暴力法是能够解决本题的,本题的数据不大,可是请不要把KMP挂上去,然后写成暴力法了.那样会误导多少后来人啊. 建议能够主要參考我的getLongestPre这个函数,看看是怎样计算最长前缀的. 怎么推…

枚举长度最短的字符串的所有子串,再与其他串匹配. #include<cstdio> #include<cstring> #include<algorithm> #include<iostream> #include<cstdlib> #include<string> #include<cmath> #include<vector> using namespace std; ; ; ); const int in…

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 recently asked…

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…

[题目链接] http://poj.org/problem?id=3450 [题目大意] 求k个字符串的最长公共子串,如果有多个答案,则输出字典序最小的. [题解] 我们对第一个串的每一个后缀和其余所有串做kmp,取匹配最小值的最大值就是答案. [代码] #include <cstring> #include <cstdio> #include <algorithm> const int N=4050,M=210; using namespace std; int nx…

题意: 给定N个字符串,寻找最长的公共字串,如果长度相同,则输出字典序最小的那个. 找其中一个字符串,枚举它的所有的字串,然后,逐个kmp比较.......相当暴力,可二分优化. #include <cstdio> #include <cmath> #include <iostream> #include <cstring> #include <string> #include <algorithm> using namespace…

http://blog.sina.com.cn/s/blog_74e20d8901010pwp.html我采用的是方法三. 注意:当长度相同时,取字典序最小的. #include <iostream> #include <stdio.h> #include <string.h> #include <algorithm> /* http://blog.sina.com.cn/s/blog_74e20d8901010pwp.html 我采用的是方法三. 注意:当…

题目链接:http://poj.org/problem?id=3450 题目分类:后缀数组 题意:求n个串的最长公共字串(输出字串) //#include<bits/stdc++.h> #include<stdio.h> #include<math.h> #include<algorithm> #include<string.h> using namespace std; #define N 200005 int wa[N],wb[N],wsf[…

<题目链接> 题目大意: 就是求k个长度为60的字符串的最长连续公共子串,2<=k<=10 限制条件: 1.  最长公共串长度小于3输出   no significant commonalities 2.  若出现等长的最长的子串,则输出字典序最小的串 解题分析: 将第一个字串的所有子串枚举出来,然后用KMP快速判断该子串是否在所有主串中出现,如果都出现过,那么就按该子串的长度和字典序,不断更新答案,直到得到最终的最优解. #include <cstdio> #incl…

Oulipo Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 26479   Accepted: 10550 Description The French author Georges Perec (1936–1982) once wrote a book, La disparition, without the letter 'e'. He was a member of the Oulipo group. A quot…

地址:http://acm.hdu.edu.cn/showproblem.php?pid=2328 题目: Corporate Identity Time Limit: 9000/3000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 1599    Accepted Submission(s): 614 Problem Description Beside other serv…

Corporate Identity Time Limit: 9000/3000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 3308    Accepted Submission(s): 1228 Problem Description Beside other services, ACM helps companies to clearly state their “cor…

http://acm.hdu.edu.cn/showproblem.php?pid=2328 Corporate Identity Time Limit: 9000/3000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 698    Accepted Submission(s): 281 Problem Description Beside other services, AC…

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 recently asked ACM for a he…

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 recently asked ACM for a he…

本文来自:http://www.cppblog.com/snowshine09/archive/2011/08/02/152272.spx 多版本的POJ分类 流传最广的一种分类: 初期: 一.基本算法: (1)枚举. (poj1753,poj2965) (2)贪心(poj1328,poj2109,poj2586) (3)递归和分治法. (4)递推. (5)构造法.(poj3295) (6)模拟法.(poj1068,poj2632,poj1573,poj2993,poj2996) 二.图算法:…

初期:一.基本算法:     (1)枚举. (poj1753,poj2965)     (2)贪心(poj1328,poj2109,poj2586)     (3)递归和分治法.     (4)递推.     (5)构造法.(poj3295)     (6)模拟法.(poj1068,poj2632,poj1573,poj2993,poj2996)二.图算法:     (1)图的深度优先遍历和广度优先遍历.     (2)最短路径算法(dijkstra,bellman-ford,floyd,hea…

poj 题目分类 按照ac的代码长度分类(主要参考最短代码和自己写的代码) 短代码:0.01K--0.50K:中短代码:0.51K--1.00K:中等代码量:1.01K--2.00K:长代码:2.01K以上. 短:1147.1163.1922.2211.2215.2229.2232.2234.2242.2245.2262.2301.2309.2313.2334.2346.2348.2350.2352.2381.2405.2406: 中短:1014.1281.1618.1928.1961.2054…

http://www.cnblogs.com/kuangbin/archive/2011/07/29/2120667.html 初期:一.基本算法:     (1)枚举. (poj1753,poj2965)     (2)贪心(poj1328,poj2109,poj2586)     (3)递归和分治法.     (4)递推.     (5)构造法.(poj3295)     (6)模拟法.(poj1068,poj2632,poj1573,poj2993,poj2996)二.图算法:     (…

Hint:补补基础... 初期:一.基本算法:     (1)枚举. (poj1753,poj2965)     (2)贪心(poj1328,poj2109,poj2586)     (3)递归和分治法.     (4)递推.     (5)构造法.(poj3295)     (6)模拟法.(poj1068,poj2632,poj1573,poj2993,poj2996)二.图算法:     (1)图的深度优先遍历和广度优先遍历.     (2)最短路径算法(dijkstra,bellman-f…

初期:一.基本算法:     (1)枚举. (poj1753,poj2965)     (2)贪心(poj1328,poj2109,poj2586)     (3)递归和分治法.     (4)递推.     (5)构造法.(poj3295)     (6)模拟法.(poj1068,poj2632,poj1573,poj2993,poj2996)二.图算法:     (1)图的深度优先遍历和广度优先遍历.     (2)最短路径算法(dijkstra,bellman-ford,floyd,hea…

题目链接:https://vjudge.net/problem/POJ-3450 Corporate Identity Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 8046   Accepted: 2710 Description Beside other services, ACM helps companies to clearly state their “corporate identity”, which i…

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 recently asked ACM for a he…

Corporate Identity Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 7662   Accepted: 2644 Description Beside other services, ACM helps companies to clearly state their "corporate identity", which includes company logo but also other…

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".…

题目链接:http://poj.org/problem?id=3450 题意:给定n个字符串,求n个字符串的最长公共子串,无解输出IDENTITY LOST,否则最长的公共子串.有多组解时输出字典序最小的解 思路:后缀数组的解法,我们把n个串都链接起来,中间用一些互不相同的且都没在原串中出现过的字符来分割开.然后求后缀数组.由于求的是最长公共子串,所以我们可以二分长度x,于是问题就转变成了是否有一个长度为x的子串在n个字符串中都出现过.判断的方式是:以height数组进行分组,height值不小…

题目链接 题意:输入N(2 <= N <= 4000)个长度不超过200的字符串,输出字典序最小的最长公共连续子串; 思路:将所有的字符串中间加上分隔符,注:分隔符只需要和输入的字符不同,且各自不同即可,没有必要是最小的字符; 连接后缀数组求解出height之后二分长度,由于height是根据sa数组建立的,所以前面符合的就是字典序最小的,直接找到就停止即可; ps: 把之前的模板简化了下,A题才是关键; #include<iostream> #include<cstdio&…

题意:求n个串的字典序最小的最长公共子串. 解法:枚举第一个串的子串,与剩下的n-1个串KMP匹配,判断是否有这样的公共子串.从大长度开始枚举,找到了就break挺快的.而且KMP的作用就是匹配子串,近乎O(n)的速度,很快. P.S.对于字符串要仔细!!! #include<cstdio> #include<cstdlib> #include<cstring> #include<iostream> using namespace std; ; int n;…

题目链接:http://poj.org/problem?id=3080 题意就是求n个长度为60的串中求最长公共子序列(长度>=3):如果有多个输出字典序最小的: 我们可以暴力求出第一个串的所有子串,然后判断是否是其他的子串即可: #include<iostream> #include<stdio.h> #include<string.h> using namespace std; ; ]; int Next[N]; void GetNext(char a[],…