PKU-1080 Human Gene Functions

原題鏈接 http://poj.org/problem?id=1080

Human Gene Functions

 

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 

 

Source Code

/*
觀察題目給出的一個最優解： 
AGTGATG 
-GTTA-G 
將其從某一處切開，如果左邊部分的分值不是最大，那麼將其進行調整，使其分值變大，
則整個解分值變大，與已知的最優矛盾。所以左邊部分的分值必是最大。
同理，右邊也是。可見滿足最優子結構的性質。考慮使用DP： 

  設兩個DNA序列分別爲s1，s2，長度分別爲len1，len2，score爲分值表。
  f[i,j]表示子串s1[1..i]和s2[1..j]的分值。考慮一個f[i,j]，我們有： 
  
	1.s1取第i個字母，s2取“-”：f[i-1,j] + score[s1[i],'-'] 
	
	2.s1取“-”，s2取第j個字母：f[i,j-1] + score['-',s2[j]] 
	  
	3.s1取第i個字母，s2取第j個字母：f[i-1,j-1] + score[s1[i],s2[j]] 
		
	即f[i,j] = max(f[i-1,j] + score[s1[i],'-'], f[i,j-1] + score['-',s2[j]], f[i-1,j-1] + score[s1[i],s2[j]]); 
		  
	然後考慮邊界條件，這道題爲i或j爲0的情況。 
	當i=j=0時，即爲f[0,0]，這是在計算f[1,1]時用到的，根據f[1,1] = f[0,0] + score[s1[i], s2[j]]，明顯有f[0,0] = 0。 
	當i=0時，即爲f[0,1..len2]，有了f[0,0]，可以用f[0,j] = f[0,j-1] + table['-',s2[j]]來計算。 
	當j=0時，即爲f[1..len1,0]，有了f[0,0]，可以用f[i,0] = f[i-1,0] + table[s1[i],'-']來計算。 
			
	至於計算順序，只要保證計算f[i,j]的時候，使用到的f[i-1,j]，f[i,j-1]，f[i-1,j-1]都計算出來了就行了。
 所謂劃分階段也就是爲了達到這個目的。這樣我們使用一個二重循環就可以了。 
*/

#include <iostream>
using namespace std;

#define Max 101
int len1,len2;
char gene1[Max];
char gene2[Max];
int dp[Max][Max];
int value[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,
				};

int change(char ch){
    if(ch=='A')
        return 0;
    else if(ch=='C')
        return 1;
    else if(ch=='G')
        return 2;
    else if(ch=='T')
        return 3;
}

int main() {
	int i, j, cases;
    while (scanf("%d", &cases) != EOF) {
		while (cases --) {
			scanf("%d %s", &len1, &gene1);
			scanf("%d %s", &len2, &gene2);

			memset(dp, 0, sizeof(dp));

			for(i = 1; i<= len1; i++)
				dp[i][0]=dp[i - 1][0] + value[change(gene1[i - 1])][4];
			for(i = 1;i<= len2;i++)
				dp[0][i]=dp[0][i - 1] + value[4][change(gene2[i - 1])];

			for (i = 1; i <= len1; i++) {
				for (j = 1; j <= len2; j++) {
					int MaxMatch = dp[i - 1][j - 1] + value[change(gene1[i - 1])][change(gene2[j - 1])];
					if (MaxMatch < dp[i - 1][j] + value[change(gene1[i - 1])][4])
						MaxMatch = dp[i - 1][j] + value[change(gene1[i - 1])][4];
					if (MaxMatch < dp[i][j - 1] + value[4][change(gene2[j - 1])])
						MaxMatch = dp[i][j - 1] + value[4][change(gene2[j - 1])];
					dp[i][j] = MaxMatch;
				}
			}
			printf("%d\n", dp[len1][len2]);
		}
     
    }
    return 0;
}