Blue Jeans
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 11542   Accepted: 4962

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

解题报告
真心不想写这解题报告。。。
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
char str[12][100],ch[100],ans[100];
int main()
{
int t,i,j,k,n,maxl,f;
cin>>t;
while(t--)
{
memset(str,0,sizeof(str));
memset(ans,0,sizeof(ans));
cin>>n;
maxl=0,f=0;
for(i=0; i<n; i++)
cin>>str[i];
for(i=0; i<58; i++)
{
for(j=3; j<=60-i; j++)
{
strncpy(ch,str[0]+i,j);
ch[j]=0;
f=0;
for(k=1; k<n; k++)
{
if(strstr(str[k],ch)==NULL)
{
f=1;
break;
}
}
if(!f)
{
if(j>maxl)
{
maxl=j;
strcpy(ans,ch);
}
else if(j==maxl)
{
if(strcmp(ch,ans)<0)
strcpy(ans,ch);
}
}
}
}
if(maxl)
cout<<ans<<endl;
else cout<<"no significant commonalities"<<endl;
}
return 0;
}

POJ训练计划3080_Blue Jeans(串处理/暴力)的更多相关文章

  1. POJ训练计划1035_Spell checker(串处理/暴力)

    Spell checker Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 18418   Accepted: 6759 De ...

  2. 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 ...

  3. 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: ...

  4. POJ 3080 Blue Jeans(串)

    题目网址:http://poj.org/problem?id=3080 思路: 以第一个DNA序列s为参考序列,开始做以下的操作. 1.将一个字母s[i]作为匹配串.(i为当前遍历到的下标) 2.遍历 ...

  5. 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 ...

  6. POJ训练计划

    POJ训练计划 Step1-500题 UVaOJ+算法竞赛入门经典+挑战编程+USACO 请见:http://acm.sdut.edu.cn/bbs/read.php?tid=5321 一.POJ训练 ...

  7. POJ 3294 n个串中至少一半的串共享的最长公共子串

    Life Forms Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 12484   Accepted: 3502 Descr ...

  8. 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 ...

  9. 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 ...

随机推荐

  1. Swift - 访问通讯录联系人(使用系统提供的通讯录交互界面)

    1,通讯录访问介绍 通讯录(或叫地址簿,电话簿)是一个数据库,里面储存了联系人的相关信息.要实现访问通讯录有如下两种方式: (1)AddressBook.framework框架 : 没有界面,通过代码 ...

  2. [置顶] ARM-Linux下WEB服务器Boa的移植、配置和运行测试

    Linux下使用的轻量级WEB服务器主要有:lighttpd.thttpd.shttpd和boa等等,而Boa是使用最为广泛的轻量级WEB服务器之一(当然,阿帕奇是世界使用排名第一的Web服务器软件) ...

  3. 节点地址的函数list_entry()原理详解

    本节中,我们继续讲解,在linux2.4内核下,如果通过一些列函数从路径名找到目标节点. 3.3.1)接下来查看chached_lookup()的代码(namei.c) [path_walk()> ...

  4. cms配置使用

    在早期完成了页面的切图之后,需要配置cms来实现小编上传数据更新页面的流程,在取得SEO的官网URL规则之后,就能开始官网在cms的基本配置了. 下面介绍cms的特点: 类别,决定内容与内容对应的路径 ...

  5. 14.8.2 Verifying File Format Compatibility 校验文件格式兼容性:

    14.8.2 Verifying File Format Compatibility 校验文件格式兼容性: 14.8.2.1 Compatibility Check When InnoDB Is St ...

  6. 被忽视的META标签之特效

    在web设计中使用js可以实现很多的页面特效,然而很多人却忽视了HTML标签中META标签的强大功效,其实meta标签也可以实现很多漂亮的页面过渡效果. META标签是HTML语言HEAD区的一个辅助 ...

  7. SQL server语句练习

    相关表: <span style="white-space:pre">create table DEPT ( <span style="white-sp ...

  8. CABasicAnimation学习Demo 包含了一些经常使用的动画效果

    个人写的一些样例: // // ViewController.m // CABasicAnimationDemo // // Created by haotian on 14-6-13. // Cop ...

  9. OpenVPN多处理之-netns容器与iptables CLUSTER

    假设还是沉湎于之前的战果以及强加的感叹,不要冥想,将其升华. 1.C还是脚本 以前,我用bash组织了复杂的iptables,ip rule等逻辑来配合OpenVPN,将其应用于差点儿全部能够想象得到 ...

  10. delphi 怎么将一个文件流转换成字符串(String到流,String到文件,相互转化)

    //from   http://kingron.myetang.com/zsfunc0d.htm (*//   标题:充分利用pascal字符串类型   说明:和PChar不同,string可以保存# ...