国产探花免费观看_亚洲丰满少妇自慰呻吟_97日韩有码在线_资源在线日韩欧美_一区二区精品毛片,辰东完美世界有声小说,欢乐颂第一季,yy玄幻小说排行榜完本

首頁 > 學院 > 開發設計 > 正文

POJ 1080 Human Gene Functions (DP)

2019-11-11 01:48:28
字體:
來源:轉載
供稿:網友

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.

img

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 file. 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

1421

題意

給定兩個基因字符串,用A,C,G,T表示其組成成分。

若兩個基因的長度不一樣,可以通過在兩個串中分別添加空格使其長度一致,當其長度一樣后,分別計算對應位置上的兩個字母的分數,并將所有的分數相加便得到兩個串的相似度分數,求最高分數。

思路

分析可知,在匹配過程中有以下幾種情況。

c[i?1]=d[j?1] ,分數為 dp[i?1][j?1]+sorce[c[i?1]][d[j?1]]c[i?1]!=d[j?1] 給i處補空格,分數 dp[i][j?1]+sorce[′?′][d[j?1]]給j處補空格,分數 dp[i?1][j]+sorce[c[i?1]][′?′]直接匹配,分數 dp[i?1][j?1]+sorce[c[i?1]][d[j?1]]

在兩字母不同下的直接匹配和相同下的匹配分數是一樣的,因此可以合并。

dp[i][j] 取這三者最大值便可。

AC 代碼

#include<iostream>#include<algorithm>#include<stdio.h>#include<string.h>#include<math.h>#include<iostream>using namespace std;#include<vector>#include<queue>int sorce[125][125];void init(){ sorce['A']['A']=5; sorce['C']['C']=5; sorce['G']['G']=5; sorce['T']['T']=5; sorce['A']['C']=sorce['C']['A']=-1; sorce['A']['G']=sorce['G']['A']=-2; sorce['A']['T']=sorce['T']['A']=-1; sorce['A']['-']=sorce['-']['A']=-3; sorce['C']['G']=sorce['G']['C']=-3; sorce['C']['T']=sorce['T']['C']=-2; sorce['C']['-']=sorce['-']['C']=-4; sorce['G']['T']=sorce['T']['G']=-2; sorce['G']['-']=sorce['-']['G']=-2; sorce['T']['-']=sorce['-']['T']=-1; sorce['-']['-']=0;}char c[115],d[115];int dp[115][115];void lcs(int lc,int ld){ memset(dp,0,sizeof(dp)); for(int i=1; i<=lc; i++) dp[i][0]=dp[i-1][0]+sorce['-'][c[i-1]]; for(int i=1; i<=ld; i++) dp[0][i]=dp[0][i-1]+sorce['-'][d[i-1]]; for(int i=1; i<=lc; i++) for(int j=1; j<=ld; j++) dp[i][j]=max(dp[i-1][j-1]+sorce[c[i-1]][d[j-1]],max(dp[i-1][j]+sorce[c[i-1]]['-'],dp[i][j-1]+sorce['-'][d[j-1]])); printf("%d/n",dp[lc][ld]);}int main(){ init(); int n; scanf("%d",&n); while(n--) { int lc,ld; cin>>lc>>c>>ld>>d; lcs(lc,ld); } return 0;}
發表評論 共有條評論
用戶名: 密碼:
驗證碼: 匿名發表
主站蜘蛛池模板: 冀州市| 武宁县| 阳新县| 福清市| 伊川县| 辽阳县| 商河县| 垣曲县| 平安县| 株洲市| 皮山县| 通化市| 社旗县| 玉林市| 会理县| 稻城县| 汝阳县| 阳新县| 贡山| 长治县| 卢龙县| 海淀区| 新安县| 荃湾区| 视频| 四子王旗| 平罗县| 福安市| 荥经县| 嘉义县| 左贡县| 抚顺县| 常德市| 嘉荫县| 木兰县| 南宫市| 绥棱县| 徐汇区| 滨海县| 福鼎市| 鹤岗市|