ZOJ1027 Human Gene Functions

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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 �C 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 �CGT--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 



7 AGTGATG 
5 GTTAG 
7 AGCTATT 
9 AGCTTTAAA


Output for the Sample Input
14 

21


描述

众所周知,人类的基因可以被视为一个序列,四个核苷酸组成,它仅仅是指由四个字母,A,C,G和T的生物学家一直在识别人类基因并确定其功能感兴趣,因为这些可以用来诊断人类疾病和他们的新药物设计。
人类基因可以通过一系列耗时的生物实验来识别,通常借助计算机程序。一旦获得一个基因序列,下一个任务就是确定它的功能。生物学家在确定一个新基因序列的功能时使用的方法之一是用新的基因查询数据库。要搜索的数据库存储了许多基因序列及其功能C。许多研究人员已经向数据库提交其基因和功能,数据库可以通过因特网自由访问。
数据库搜索将返回与查询基因相似的数据库中的基因序列列表。生物学家认为序列相似性通常意味着功能相似性。因此,新基因的功能可能是该列表中的基因所具有的功能之一。为了准确确定哪一个是正确的,需要进行另一系列的生物学实验。
你的工作是制定一个程序,比较两个基因,并确定它们的相似性,如下所述。如果你能提供一个有效的数据库,你的程序可能被用作数据库搜索的一部分。
给出两基因agtgatg和gttag,他们是多么的相似?测量两个基因相似性的方法之一称为比对。在一个对齐中,如果需要,在适当的位置插入空格,使它们等长,并根据得分矩阵对所得的基因进行打分。
例如,一个空间插入agtgatg导致agtgat-g,和三位插入gttag导致�CGT——标签。空间由减号(-)表示。这两个基因的长度相等。这两个字符串是对齐的:
agtgat-g
gt标记
在这个对齐中,有四个匹配,即g在第二个位置,t在第三,t在第六,g在第八。每一对对齐的字符根据下面的得分矩阵分配一个分数。
*表示不允许进行空间空间匹配。以上对齐的分数是(- 3)+ 5 + 5 +(- 2)+(- 3)+ 5 +(- 3)+ 5 = 9。
当然,许多其他的对齐方式都是可能的。如下所示(不同的空格被插入不同的位置):
agtgatg
- gtta-g
这个对齐给出了一个分数(- 3)+ 5 + 5 +(- 2)+ 5 +(- 1)+ 5 = 14。所以,这一个比上一个好。事实上,这一个是最佳的,因为没有其他的对齐可以有更高的分数。因此,这两个基因的相似性是14。
输入
输入由t测试用例组成。测试用例的数量(t在输入的第一行中给出)。每个测试用例由两行组成:每行包含一个整数,一个基因的长度,然后是一个基因序列。每个基因序列的长度至少是一个,不超过100个。
输出
输出应该打印每个测试用例的相似性,每行一个。
样本输入

7 agtgatg
5 gttag
7 agctatt
9 agctttaaa
示例输入的输出
十四

二十一


#include#include#include#include#include#includeusing namespace std;int g[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,-1,0}};char str1[105];char str2[105];map m;int cas;int dp[105][105];int main(){int len1,len2;scanf("%d",&cas);m['A'] = 0;m['C'] = 1;m['G'] = 2;m['T'] = 3;m['-'] = 4;   while(cas--)   {    scanf("%d%s",&len1, str1);    scanf("%d%s",&len2, str2);    //边界条件    dp[0][0] = 0;    for(int i=1;i<=len2;i++)    dp[0][i] = dp[0][i-1] + g[4][m[str2[i-1]]];    for(int i=1;i<=len1;i++)    dp[i][0] = dp[i-1][0] + g[m[str1[i-1]]][4];    int m1,m2,m3;    for(int i=1;i<=len1;i++)    for(int j=1;j<=len2;j++)                            //动态规划    {        m1 = dp[i-1][j]+g[m[str1[i-1]]][4];        m2 = dp[i][j-1]+g[4][m[str2[j-1]]];        m3 = dp[i-1][j-1]+g[m[str1[i-1]]][m[str2[j-1]]];        dp[i][j]=max(m1,max(m2,m3));    }    printf("%d\n",dp[len1][len2]);   }}/*#include #include #include using namespace std;const int maxn = 110;const int score[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, 0 }                        };char str1[maxn], str2[maxn];int gene[maxn][maxn], Map[128];inline int max3(int a, int b, int c){    return max(max(a, b), c);}int main(){    int T, len1, len2;    Map['A'] = 0; Map['C'] = 1; Map['G'] = 2; Map['T'] = 3; Map['-'] = 4;    scanf("%d", &T);    while(T--) {        scanf("%d%s%d%s", &len1, str1, &len2, str2);        gene[0][0] = 0;        for(int i=1; i<=len2; i++)            gene[0][i] = gene[0][i-1] + score[4][Map[str2[i-1]]];        for(int i=1; i<=len1; i++)            gene[i][0] = gene[i-1][0] + score[Map[str1[i-1]]][4];        for(int i=1; i<=len1; i++)            for(int j=1; j<=len2; j++) {                gene[i][j] = max3(gene[i-1][j] + score[Map[str1[i-1]]][4],                                gene[i][j-1] + score[4][Map[str2[j-1]]],                                gene[i-1][j-1] + score[Map[str1[i-1]]][Map[str2[j-1]]]);            }        printf("%d\n", gene[len1][len2]);    }    return 0;}  *//*#include#include#include#include#include#includeusing namespace std;int g[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,-1,0}};char str1[105];char str2[105];map m;int cas;int dp[105][105];//前i个数和前j个数已匹配的情况int main(){  // freopen("input.txt","r",stdin);   int len1,len2;   scanf("%d",&cas);   m['A']=0;   m['C']=1;   m['G']=2;   m['T']=3;   m['-']=4;   while(cas--)   {     scanf("%d%s",&len1,str1);     scanf("%d%s",&len2,str2);     dp[0][0]=0;     for(int i=1;i<=len1;i++)        dp[i][0]=dp[i-1][0]+g[m[str1[i-1]]][4];     for(int i=1;i<=len2;i++)        dp[0][i]=dp[0][i-1]+g[4][m[str2[i-1]]];     for(int i=1;i<=len1;i++)        for(int j=1;j<=len2;j++)     {         int t1=dp[i-1][j-1]+g[m[str1[i-1]]][m[str2[j-1]]];         int t2=dp[i-1][j]+g[m[str1[i-1]]][4];         int t3=dp[i][j-1]+g[4][m[str2[j-1]]];         dp[i][j]=max(t1,max(t2,t3));     }     printf("%d\n",dp[len1][len2]);   }}*/

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