博客
关于我
强烈建议你试试无所不能的chatGPT,快点击我
POJ3080-Blue Jeans
阅读量:6582 次
发布时间:2019-06-24

本文共 3238 字,大约阅读时间需要 10 分钟。

 

题目链接:

 

 

Blue Jeans

Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 20430   Accepted: 9053

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

32GATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA3GATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATAGATACTAGATACTAGATACTAGATACTAAAGGAAAGGGAAAAGGGGAAAAAGGGGGAAAAGATACCAGATACCAGATACCAGATACCAAAGGAAAGGGAAAAGGGGAAAAAGGGGGAAAA3CATCATCATCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCACATCATCATAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAACATCATCATTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTT

Sample Output

no significant commonalitiesAGATACCATCATCAT

题目大意:找3个所给字符串的最长公共子序列,(长度大于等于3)

思路:拿第一个字符串,暴力取字串,长度递增,去其他字符串里面找。

AC代码:

#include
#include
#include
#include
#include
using namespace std;string str[15];string temp;vector
t;vector
ans;bool cmp(string a, string b) {//优先顺序 if(a.size() == b.size()) return a < b; else return a.size() > b.size();} int main() { int n, m; while(cin >> n) { while(n--) { cin >> m; for(int i = 0; i < m; i++) {//存储 cin >> str[i]; } ans.clear(); for(int j = 3; j <= 60; j++) {//最短3,最长60 for(int i = 0; i <= str[0].size() - j; i++) {//注意等号 int flag = 0; temp = str[0].substr(i, j);//从i开始,取长度为j的字串 for(int k = 1; k < m; k++) {//去其他字符串里面找 if(str[k].find(temp) == string::npos) { flag = 1; break; } } if(flag == 0) ans.push_back(temp);//合法的 } } sort(ans.begin(), ans.end(), cmp);//第一个就是 if(ans.size() != 0) if(ans[0].size() >= 3) cout << ans[0] << endl; else cout << "no significant commonalities" << endl; else cout << "no significant commonalities" << endl; } }}

 

 

 

转载于:https://www.cnblogs.com/ACMerszl/p/9572974.html

你可能感兴趣的文章
linux中设置静态ip
查看>>
Android学习——HorizontalScollview水平滚动控件
查看>>
我的友情链接
查看>>
Python多线程ping检测机器存活(跨平台版)
查看>>
CIsco 3845 静态DNS条目(ip host)自动消失
查看>>
f2fs系列 之三:目录结构
查看>>
DBMS的工作模式
查看>>
aix文件系统在线扩容
查看>>
swoole教程第一节:进程管理模块(Process)-中(消息队列)
查看>>
java.lang.IndexOutOfBoundsException: Invalid index 0, size is 0
查看>>
有意思的文章的链接
查看>>
Entity Framework技术系列之6:数据绑定
查看>>
2019-05-13 Java学习日记之循环语句&方法
查看>>
MySQL 优化工具 MySQLTuner
查看>>
读取文件并替换内容
查看>>
mysql5.6延迟复制
查看>>
LInux 常用命令
查看>>
ASP 格式化时间函数
查看>>
我的友情链接
查看>>
更换mysql数据目录后出现: socket '/var/lib/mysql/mysql.sock' (2)
查看>>