【poj3080 Blue Jeans【KMP】【暴力】】的更多相关文章

POJ3080 - Blue Jeans(KMP+二分)

题目大意 求N个字符串的最长公共字串 题解 和POJ1226做法一样...注意是字典序最小的...WA了一次 代码: #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> using namespace std; #define MAXN 65 char p[MAXN],T[MAXN][MAXN]; int f[MAXN]; void getfail(cha…

POJ3080 Blue Jeans —— 暴力枚举 + KMP / strstr()

题目链接:https://vjudge.net/problem/POJ-3080 Blue Jeans Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 19152   Accepted: 8524 Description The Genographic Project is a research partnership between IBM and The National Geographic Society that…

poj3080 Blue Jeans【KMP】【暴力】

Blue Jeans Time Limit: 1000MS   Memory Limit: 65536K Total Submissions:21746   Accepted: 9653 Description The Genographic Project is a research partnership between IBM and The National Geographic Society that is analyzing DNA from hundreds of thousan…

POJ3080——Blue Jeans(暴力+字符串匹配)

Blue Jeans DescriptionThe Genographic Project is a research partnership between IBM and The National Geographic Society that is analyzing DNA from hundreds of thousands of contributors to map how the Earth was populated. As an IBM researcher, you hav…

POJ 3080 Blue Jeans(Java暴力)

Blue Jeans [题目链接]Blue Jeans [题目类型]Java暴力 &题意: 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 规定: 1. 最长公共串长度小于3不输出 2. 若出现等长的最长的子串,则输出字典序最小的串 &题解: 这个我刚开始用c++写的,真的是恶心到我了,啥都要自己实现,所以突然就想到用Java试试,结果还真的挺简单.思路就是在第1个串中找出所有子串,之后去下面找就行了. String.compareTo(Sting )是字符串…

kuangbin专题十六 KMP&&扩展KMP POJ3080 Blue Jeans

The Genographic Project is a research partnership between IBM and The National Geographic Society that is analyzing DNA from hundreds of thousands of contributors to map how the Earth was populated. As an IBM researcher, you have been tasked with wri…

poj3080 Blue Jeans(暴枚+kmp)

Description The Genographic Project is a research partnership between IBM and The National Geographic Society that is analyzing DNA from hundreds of thousands of contributors to map how the Earth was populated.  As an IBM researcher, you have been ta…

POJ3080 Blue Jeans 题解 KMP算法

题目链接:http://poj.org/problem?id=3080 题目大意:给你N个长度为60的字符串(N<=10),求他们的最长公共子串(长度>=3). 题目分析:KMP字符串匹配基础题.直接枚举第1个字符串的所有子串,判断这个子串是否出现在另外N-1个串中. 实现代码如下: #include <cstdio> #include <string> #include <iostream> #include <algorithm> #incl…

POJ-3080-Blue jeans(KMP, 暴力)

链接: https://vjudge.net/problem/POJ-3080#author=alexandleo 题意: 给你一些字符串,让你找出最长的公共子串. 思路: 暴力枚举第一个串的子串,挨个匹配接下来的所有串. 找出最长子串中, 字典序最小的. 代码: #include <iostream> #include <cstdio> #include <cstring> #include <vector> //#include <memory.h…

Blue Jeans---poj3080(kmp+暴力求子串)

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[],…