hdu1080---Human Gene Functions
发布日期:2022-02-02 02:58:12 浏览次数:22 分类:技术文章

本文共 4857 字,大约阅读时间需要 16 分钟。

Problem Description
It is well known that a human gene can be considered as a sequence, consisting of four nucleotides, which are simply denoted by four letters, A, C, G, and T. Biologists have been interested in identifying human genes and determining their functions, because these can be used to diagnose human diseases and to design new drugs for them.
A human gene can be identified through a series of time-consuming biological experiments, often with the help of computer programs. Once a sequence of a gene is obtained, the next job is to determine its function. One of the methods for biologists to use in determining the function of a new gene sequence that they have just identified is to search a database with the new gene as a query. The database to be searched stores many gene sequences and their functions – many researchers have been submitting their genes and functions to the database and the database is freely accessible through the Internet.
A database search will return a list of gene sequences from the database that are similar to the query gene. Biologists assume that sequence similarity often implies functional similarity. So, the function of the new gene might be one of the functions that the genes from the list have. To exactly determine which one is the right one another series of biological experiments will be needed.
Your job is to make a program that compares two genes and determines their similarity as explained below. Your program may be used as a part of the database search if you can provide an efficient one.
Given two genes AGTGATG and GTTAG, how similar are they? One of the methods to measure the similarity of two genes is called alignment. In an alignment, spaces are inserted, if necessary, in appropriate positions of the genes to make them equally long and score the resulting genes according to a scoring matrix.
For example, one space is inserted into AGTGATG to result in AGTGAT-G, and three spaces are inserted into GTTAG to result in –GT--TAG. A space is denoted by a minus sign (-). The two genes are now of equal length. These two strings are aligned:
AGTGAT-G
-GT--TAG
In this alignment, there are four matches, namely, G in the second position, T in the third, T in the sixth, and G in the eighth. Each pair of aligned characters is assigned a score according to the following scoring matrix.
* denotes that a space-space match is not allowed. The score of the alignment above is (-3)+5+5+(-2)+(-3)+5+(-3)+5=9.
Of course, many other alignments are possible. One is shown below (a different number of spaces are inserted into different positions):
AGTGATG
-GTTA-G
This alignment gives a score of (-3)+5+5+(-2)+5+(-1) +5=14. So, this one is better than the previous one. As a matter of fact, this one is optimal since no other alignment can have a higher score. So, it is said that the similarity of the two genes is 14.
 
Input
The input consists of T test cases. The number of test cases ) (T is given in the first line of the input. Each test case consists of two lines: each line contains an integer, the length of a gene, followed by a gene sequence. The length of each gene sequence is at least one and does not exceed 100.
 
Output
The output should print the similarity of each test case, one per line.
 
Sample Input
2 7 AGTGATG 5 GTTAG 7 AGCTATT 9 AGCTTTAAA
 
Sample Output
14 21
题目大意:

给定两个字符串a,b;其中两个字符串中的字母两两对应都有一个权值,并且两个字符串可以任意添加空格,使得有更多的字符可以对应,例如Given two genes AGTGATG and GTTAG,可以写成

AGTGAT-G

-GT--TAG 这样的最终对应权值是The score of the alignment above is (-3)+5+5+(-2)+(-3)+5+(-3)+5=9.

也可以写成

AGTGATG

-GTTA-G 这样的对应权值之和是(-3)+5+5+(-2)+5+(-1) +5=14,因为空格的加入,可以有很多种不同的权值和,要求的是最大的权值和

代码实现:

#include
#include
using namespace std;int a[5][5]= {5,-1,-2,-1,-3, -1,5,-3,-2,-4, -2,-3,5,-2,-2, -1,-2,-2,5,-1, -3,-4,-2,-1,100000 };char str1[101],str2[101];int dp[101][101];int find(char a){ if(a=='A') return 0; if(a=='C') return 1; if(a=='G') return 2; if(a=='T') return 3; if(a=='-') return 4;}int main(){ int n,len1,len2; cin>>n; while(n--) { cin>>len1; for(int i=1;i<=len1;i++) { cin>>str1[i]; } cin>>len2; for(int j=1;j<=len2;j++) { cin>>str2[j]; } int len=max(len1,len2); memset(dp,0,sizeof(dp)); for(int i=1; i<=len; i++) { dp[0][i]=dp[0][i-1]+a[find('-')][find(str2[i])]; dp[i][0]=dp[i-1][0]+a[find(str1[i])][find('-')]; } for(int i=1; i<=len1; i++) { for(int j=1; j<=len2; j++) { dp[i][j]=max(dp[i-1][j-1]+a[find(str1[i])][find(str2[j])],max(dp[i-1][j]+a[find(str1[i])][find('-')],dp[i][j-1]+a[find('-')][find(str2[j])])); } } cout<
<

转载地址:https://blog.csdn.net/u010368749/article/details/20391733 如侵犯您的版权,请留言回复原文章的地址,我们会给您删除此文章,给您带来不便请您谅解!

上一篇:uva1133---Vacation
下一篇:hdu1423---Greatest Common Increasing Subsequence(最长公共上升子序列)

发表评论

最新留言

路过,博主的博客真漂亮。。
[***.116.15.85]2024年03月03日 09时35分25秒

关于作者

    喝酒易醉,品茶养心,人生如梦,品茶悟道,何以解忧?唯有杜康!
-- 愿君每日到此一游!

推荐文章

php语言冒泡法,PHP 冒泡排序法 2019-04-21
php如何数组去重复,PHP如何去除数组重复元素? 2019-04-21
java转换ab的值,查看新闻/公告--[整理]Java将AB1234形式的16进制字符串转换为10进制数值,考虑字节序的影响.... 2019-04-21
ui php h5,画出自己的UI组件的详情 2019-04-21
linux服务文件编写,linux编写systemd下服务脚本 2019-04-21
hdfs linux 目录是否存在,Linux中判断hdfs文件是否存在 2019-04-21
linux学习需要什么基础,学linux需要什么基础? 2019-04-21
linux vim编辑kconfig 无法wq,Linux-4.9.2内核在mini2440上的移植(三)——编译环境测试... 2019-04-21
高斯勒让德在c语言中的程序,c语言:用递归方法编写程序,求n阶勒让德多项式的值... 2019-04-21
c语言单片机电子时钟,新人求个51单片机的电子时钟汇编语言(C语言的还没学到)... 2019-04-21
c++语言文件流,C++文件流 2019-04-21
android 动态毛玻璃,Android毛玻璃背景效果简单实现代码 2019-04-21
android 按钮提示,的Android按钮工具提示 2019-04-21
iphone通讯录 android,3个方法,教你如何快速而又有效的将联系人从iPhone转移到安卓... 2019-04-21
android horizontalscrollview 滑动事件,ScrollView的滑动监听(以HorizontalScrollView为例) 2019-04-21
win7自定义html为桌面,Win7系统自定义桌面主题的方法 2019-04-21
单系统 台电x80pro_台电x80 pro (ID:E3E6)安装remix OS系统教程整理 2019-04-21
linux存储pdf伟岸_python的reportlab库介绍、制作pdf和作图 2019-04-21
安徽信息技术初中会考上机考试模拟_2020年中小学寒假、考试时间定下了! 2019-04-21
ubuntu 退出anaconda环境_从零开始深度学习第15讲:ubuntu16.04 下深度学习开发环境搭建与配置... 2019-04-21