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Time Limit: 1000MS |
Memory Limit: 10000K |
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Total Submissions: 20750 |
Accepted: 11563 |
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 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
14
21
package POJ;
import java.util.HashMap;
import java.util.Map;
import java.util.Scanner;
public class P1080 {
/*
A C G T -
0 1 2 3 4
A C G T -
A 5 -1 -2 -1 -3
C -1 5 -3 -2 -4
G -2 -3 5 -2 -2
T -1 -2 -2 5 -1
- -3 -4 -2 -1 *
*/
public static int getMax(int a,int b,int c){
return Math.max(Math.max(a,b),Math.max(a,c));
}
public static void main(String args[]) {
int[][] score={ {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} };
Map<Character,Integer> map=new HashMap<Character, Integer>();
map.put('A',0);
map.put('C',1);
map.put('G',2);
map.put('T',3);
map.put('-',4);
Scanner cin=new Scanner(System.in);
int pairs=Integer.valueOf(cin.nextLine().substring(0,1));//基因对数。主要是对数后面跟了个空格，不能直接转为数字
for(int k=0;k<pairs;k++)
{
//int similarity=0;
String str1=cin.nextLine();
int spaceIndex1=str1.indexOf(" ");//index是第一个空格的index
int len1=Integer.valueOf(str1.substring(0,spaceIndex1));
String seq1=str1.substring(spaceIndex1+1,spaceIndex1+len1+1);
String str2=cin.nextLine();
int spaceIndex2=str1.indexOf(" ");
int len2=Integer.valueOf(str2.substring(0,spaceIndex2));
String seq2=str2.substring(spaceIndex2+1,spaceIndex2+len2+1);
int[][] dp=new int[len1+1][len2+1];//dp[i][j]用来表示字符串s1 1~i的和字符串序列s2 1~j的最大相似度
/*
三种情况：
Ai和Bj匹配
Ai和空格匹配，A_(i-1)再和B_j匹配
Bj和空格匹配，A_i再和B_(j-1)匹配
*/
/*
dp数组边缘初始化
dp[i][j] = max{dp[i-1][j-1]+value(a[i],b]j],dp[i][j-1]+value(-,b[j]), dp[i-1][j]+value(a[i],-)}
*/
dp[0][0] = 0;
for(int i=0;i<len1;i++)//字符串下标0~len1-1。dp下标0~len1
{
dp[i+1][0] = dp[i][0] + score[map.get(seq1.charAt(i))][map.get('-')];
}
for(int j=0;j<len2;j++)
{
/*System.out.println(dp[0][j]);
System.out.println(seq2.charAt(j));
System.out.println(map.get(seq2.charAt(j)));
System.out.println(score[map.get('-')][map.get(seq2.charAt(j))]);
System.out.println(dp[0][j+1]);*/
dp[0][j+1] = dp[0][j] + score[map.get('-')][map.get(seq2.charAt(j))];
}
for(int i=1;i<=len1;i++){
for(int j=1;j<=len2;j++){
/*
dp下标与seq下标不同
大1
*/
dp[i][j]=getMax(dp[i-1][j-1] + score[map.get(seq1.charAt(i-1))][map.get(seq2.charAt(j-1))],
dp[i-1][j] +score[map.get(seq1.charAt(i-1))][map.get('-')] ,
dp[i][j-1] + score[map.get('-')][map.get(seq2.charAt(j-1))]);
}
}
/* for(int i=0;i<=len1;i++){
for(int j=0;j<=len2;j++){
*//*
dp下标与seq下标不同
大1
*//*
System.out.print(dp[i][j]+" ");
}
System.out.println();
}*/
System.out.println(dp[len1][len2]);
}
}
}