Time Limit: 1000MS
Memory Limit: 65536K

Total Submissions: 19026
Accepted: 8466

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 tasked with writing a program that will find commonalities amongst given snippets of DNA that can be correlated with individual survey information to identify new genetic markers.
A DNA base sequence is noted by listing the nitrogen bases in the order in which they are found in the molecule. There are four bases: adenine (A), thymine (T), guanine (G), and cytosine (C). A 6-base DNA sequence could be represented as TAGACC.
Given a set of DNA base sequences, determine the longest series of bases that occurs in all of the sequences.

Input

Input to this problem will begin with a line containing a single integer n indicating the number of datasets. Each dataset consists of the following components:

  • A single positive integer m (2 <= m <= 10) indicating the number of base sequences in this dataset.
  • m lines each containing a single base sequence consisting of 60 bases.

Output

For each dataset in the input, output the longest base subsequence common to all of the given base sequences. If the longest common subsequence is less than three bases in length, display the string "no significant commonalities" instead. If multiple subsequences of the same longest length exist, output only the subsequence that comes first in alphabetical order.

Sample Input

3
2
GATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATA
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
3
GATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATA
GATACTAGATACTAGATACTAGATACTAAAGGAAAGGGAAAAGGGGAAAAAGGGGGAAAA
GATACCAGATACCAGATACCAGATACCAAAGGAAAGGGAAAAGGGGAAAAAGGGGGAAAA
3
CATCATCATCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC
ACATCATCATAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
AACATCATCATTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTT

Sample Output

no significant commonalities
AGATAC
CATCATCAT

Source

South Central USA 2006

【翻译】给出一些字符串,找出最长公共连续子串,并且输出字典序最小的那个,如果最优解长度小于3,输出no significant commonalities。

题解:

       ①串长度不超过60,因此考虑暴力枚举第一个串的子串。

       ②在上文的情况下,将当前枚举的子串与剩下的字符串匹配就可以了。

       ③为了找到字典序最小,这里直接用STL-string操作比较方便。

#include<string>
#include<stdio.h>
#include<cstring>
#include<iostream>
#define go(i,a,b) for(int i=a;i<=b;i++)
#define ro(i,a,b) for(int i=a;i>=b;i--)
using namespace std;
const int N=100;string ans,tmp;
int t,n,m,C,len,ok,OK,f[N];
char s[15][N],T[N],P[N];
bool KMP_Matching()
{
int j;f[0]=f[1]=0;
go(i,1,m-1){j=f[i];while(j&&P[i]!=P[j])j=f[j];f[i+1]=P[i]==P[j]?j+1:0;}j=0;
go(i,0,n-1){while(j&&P[j]!=T[i])j=f[j];if((j+=P[j]==T[i])==m)return 1;}
return 0;
}
int main()
{
scanf("%d",&C);
while(C--&&scanf("%d",&t))
{
go(i,1,t)scanf("%s",s[i]);
ans="";len=strlen(s[1]);
ro(k,len,1)
{
OK=0;
go(i,0,len-k)
{
ok=1;m=0;
go(u,i,i+k-1)P[m++]=s[1][u];
go(j,2,t)
{
n=strlen(s[j]);
go(u,0,n-1)T[u]=s[j][u];
if(!(ok&=KMP_Matching()))break;
}
if(ok)
{
OK=1;tmp.clear();go(u,0,m-1)tmp+=P[u];
if(ans==""||tmp<ans)ans=tmp;
}
}
if(OK)
{
if(ans.length()>=3)cout<<ans<<endl;
else puts("no significant commonalities");goto Judy;
}
}
puts("no significant commonalities");Judy:;
}
return 0;
}//Paul_Guderian

谁能告诉我那奔腾的迷惘与骄傲,

是否就是我心底永隔一世的河流。——————汪峰《河流》

【POJ 3080 Blue Jeans】的更多相关文章

  1. 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 ...

  2. 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 ...

  3. poj 3080 Blue Jeans

    点击打开链接 Blue Jeans Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 10243   Accepted: 434 ...

  4. 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: ...

  5. 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 Descr ...

  6. poj 3080 Blue Jeans【字符串处理+ 亮点是:字符串函数的使用】

    题目:http://poj.org/problem?id=3080 Sample Input 3 2 GATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCA ...

  7. POJ - 3080 Blue Jeans 【KMP+暴力】(最大公共字串)

    <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 co ...

  8. POJ 3080 Blue Jeans(后缀数组+二分答案)

    [题目链接] http://poj.org/problem?id=3080 [题目大意] 求k个串的最长公共子串,如果存在多个则输出字典序最小,如果长度小于3则判断查找失败. [题解] 将所有字符串通 ...

  9. 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 ...

随机推荐

  1. .Net Core爬虫爬取妹子网图片

    现在网上大把的Python的爬虫教程,很少看见有用C#写的,正好新出的.Net Core可以很方便的部署到Linux上,就用妹子图做示范写个小爬虫 在C#下有个很方便的类库 HtmlAgilityPa ...

  2. Python学习之登陆认证

    需求: 让用户输入用户名密码 认证成功后显示欢迎信息 输错三次后退出程序 可以支持多个用户登录 (提示,通过列表存多个账户信息) 用户3次认证失败后,退出程序,再次启动程序尝试登录时,还是锁定状态(提 ...

  3. select值改变

    改变select的值,然后执行一个方法.可以用chang: $("#select").change(function(){ //要执行的内容 });

  4. [BZOJ3172 ][Tjoi2013]单词(AC自动机)

    Description 不稳定的传送门 某人读论文,一篇论文是由许多单词组成.但他发现一个单词会在论文中出现很多次,现在想知道每个单词分别在论文中出现多少次.单词个数<=200,单词总长度< ...

  5. java实时监听日志写入kafka

    目的 实时监听某目录下的日志文件,如有新文件切换到新文件,并同步写入kafka,同时记录日志文件的行位置,以应对进程异常退出,能从上次的文件位置开始读取(考虑到效率,这里是每100条记一次,可调整) ...

  6. mysql in和exists性能比较和使用【转】

    exists对外表用loop逐条查询,每次查询都会查看exists的条件语句,当 exists里的条件语句能够返回记录行时(无论记录行是的多少,只要能返回),条件就为真,返回当前loop到的这条记录, ...

  7. .Net Mvc 4 Route路由

    1.前言 在创建一个MVC项目后就可以,在App_Start文件下的RouteConfig.cs里面就可以定义我们的路由规则,其中已经有默认的路由规则了,routes.IgnoreRoute是让路由规 ...

  8. 基于CSS多列实现瀑布流

    <!DOCTYPE html> <html lang="en"> <head> <meta charset="UTF-8&quo ...

  9. android gridview 停止滚动

    http://blog.csdn.net/yaphetzhao/article/details/50544105 参考上面的博客,关键代码我就贴出来吧: public void stopGridVie ...

  10. React + webpack 快速搭建开发环境

    因网上大多React + webpack快速搭建的运行不起来,便自行写了一个.在搭建开发环境的前需安装nodejs,npm. 新建一个工作目录,比如叫reactdome,在reactdome目录中运行 ...