POJ3080 Blue Jeans

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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(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…

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(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…

题目链接. 题目大意: 给定n个字符串,找出最长相同且长度大于3的子串,如果存在多个,找出字典序最小的. 分析: 直接枚举(暴搜). 对于s[0]的每一个子串,判断是否在其它n-1个字符串中都存在. #include <iostream> #include <cstdio> #include <string> #include <map> using namespace std; #define N 60 ], str, str_max; int main(…

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…

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 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

点击打开链接 Blue Jeans Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 10243 Accepted: 4347 Description The Genographic Project is a research partnership between IBM and The National Geographic Society that is analyzing DNA from hundreds of…

POJ 3080 Blue Jeans (字符串处理暴力枚举）

Blue Jeans Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 21078 Accepted: 9340 Description The Genographic Project is a research partnership between IBM and The National Geographic Society that is analyzing DNA from hundred…

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…

（字符串 KMP）Blue Jeans -- POJ -- 3080:

链接: http://poj.org/problem?id=3080 http://acm.hust.edu.cn/vjudge/contest/view.action?cid=88230#problem/E (密码0817) Blue Jeans Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 14544 Accepted: 6478 Description The Genographic Project is a…

POJ 3080 Blue Jeans 找最长公共子串（暴力模拟+KMP匹配）

Blue Jeans Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 20966 Accepted: 9279 Description The Genographic Project is a research partnership between IBM and The National Geographic Society that is analyzing DNA from hundreds of thousa…

POJ Blue Jeans [枚举+KMP]

传送门 F - Blue Jeans Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u Submit Status Description The Genographic Project is a research partnership between IBM and The National Geographic Society that is analyzing DNA from hun…

POJ 3080 Blue Jeans (求最长公共字符串）

POJ 3080 Blue Jeans (求最长公共字符串) 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 IB…

POJ3080：Blue Jeans

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 tas…

Blue Jeans[poj3080]题解

题目 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 bee…

POJ-3080 Blue Jeans---字符串+暴力

题目链接: https://vjudge.net/problem/POJ-3080 题目大意: 找最长的公共字串(长度>=3),长度相同就找字典序最小的解题思路: 枚举第一个串的所以子串,处理出其他串的所有子串,然后set查找,更新ans #include<iostream> #include<algorithm> #include<set> #include<string> using namespace std; int T; ]; int ma…

POJ 3080 Blue Jeans、POJ 3461 Oulipo——KMP应用

题目:POJ3080 http://poj.org/problem?id=3080 题意:对于输入的文本串,输出最长的公共子串,如果长度相同,输出字典序最小的. 这题数据量很小,用暴力也是16ms,用后缀数组可以到0ms,但我不会XD. 暴力: #include<cstdio> #include<cstring> using namespace std; ][],ans[]; int main(){ int T,n; scanf("%d", &T); w…

Match:Blue Jeans(POJ 3080)

DNA序列题目大意:给你m串字符串,要你找最长的相同的连续字串这题暴力kmp即可,注意要按字典序排序,同时,是len<3才输出no significant commonalities #include <iostream> #include <functional> #include <algorithm> #include <string.h> #define MAX 60 using namespace std; typedef char* _…

【poj 3080】Blue Jeans（字符串--KMP+暴力枚举+剪枝）

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;…

poj 3080 Blue Jeans 解题报告

题目链接:http://poj.org/problem?id=3080 该题属于字符串处理中的串模式匹配问题.题目要求我们:给出一个DNA碱基序列,输出最长的相同的碱基子序列.(保证在所有的序列中都有出现) 这里采用了Brute Force算法(由于碱基序列的串长仅为60,规模比较小),这是模式匹配的一种最简单的做法. 设: 最长公共字串为ans,其长度为maxlen. m个碱基序列为p[0]...p[m-1].由于公共子序列是每个碱基序列的子串,因此不妨枚举p[0]的每一个可能的子串s.以s为…

POJ 3080 Blue Jeans （多个字符串的最长公共序列，暴力比较）

题意:给出m个字符串,找出其中的最长公共子序列,如果相同长度的有多个,输出按字母排序中的第一个. 思路:数据小,因此枚举第一个字符串的所有子字符串s,再一个个比较,是否为其它字符串的字串.判断是否为字串的时候,将s的字符依次与其他字符串的字符比较. #include <iostream> #include <stdio.h> #include <string.h> #include <stack> #include <algorithm> usi…

POJ 3080 (字符串水题) Blue Jeans

题意: 找出这些串中最长的公共子串(长度≥3),如果长度相同输出字典序最小的那个. 分析: 用库函数strstr直接查找就好了,用KMP反而是杀鸡用牛刀. #include <cstdio> #include <cstring> ][], sub[]; ], l; int cmp(int p1, int p2) { ; i < l; ++i) ][p1 + l] > a[][p2 + l]) return p2; return p1; } int main(void)…

poj 3080 Blue Jeans（水题暴搜）

题目:http://poj.org/problem?id=3080 水题,暴搜 #include <iostream> #include<cstdio> #include<cstring> #include<cstdlib> #include<stack> #include<queue> #include<iomanip> #include<cmath> #include<map> #include…

POJ 3080 Blue Jeans （KMP）

求出公共子序列要求最长字典序最小枚举第一串的所有子串然后对每一个串做KMP.找到目标子串学会了 strncpy函数的使用我已可入灵魂 #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> using namespace std; char str[20][70]; char tmp[70],ans[70]; int f[70];…

Blue Jeans