POJ 1080 Human Gene Functions(DP LCS)

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 

题目大意：给定两个字符串在两串中可以在任意位置上添加空格，并使兩串对齐，使得两字符串对应值相加后最大。

思路：此题和LCS，类似主要就是状态方程的问题因为任何一个字母都有可能和‘ ’组合，所以先初始化一个表格，也在程序中进行初始化，即每行的第一个数与‘  ’组合的情况故有dp[i][0]=dp[i-1][0]+match[a[i]]['  '](类似的每列也是这么初始化)。

转移方程：dp[i][j]=max(  dp[i-1][j]+match[a[i]]['  '] ,dp[i][j-1]+match['  '][b[j]] ,dp[i][j],dp[i-1][j-1]+match[a[i]][b[j]] ).

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#define LL long long
#define inf 0x3f3f3f3f
using namespace std;
int dp[500][500],match[100][100];
char a[125],b[125];
void ini()//该图表弱复制：http://www.cnblogs.com/zhourongqing/archive/2012/08/09/2629840.html
{//感谢大神。。
    match['A']['A']=match['C']['C']=match['G']['G']=match['T']['T']=5;
    match['A']['C']=match['C']['A']=match['A']['T']=match['T']['A']=-1;
    match[' ']['T']=match['T'][' ']=-1;
    match['A']['G']=match['G']['A']=match['C']['T']=match['T']['C']=-2;
    match['G']['T']=match['T']['G']=match['G'][' ']=match[' ']['G']=-2;
    match['A'][' ']=match[' ']['A']=match['C']['G']=match['G']['C']=-3;
    match['C'][' ']=match[' ']['C']=-4;
}
int main()
{
    int n,m,x,k,i,j;
    int cla;
    ini();
    scanf("%d",&cla);
    while(cla--)
    {
        dp[0][0]=0;
        scanf("%d%s",&n,a+1);
        scanf("%d%s",&m,b+1);
        for(i=1; i<=n; i++)
            dp[i][0]=dp[i-1][0]+match[a[i]][' '];
        for(i=1; i<=m; i++)
            dp[0][i]=dp[0][i-1]+match[' '][b[i]];
        for(i=1;i<=n;i++)
        {
            for(j=1;j<=m;j++)
            {
                dp[i][j]=max(dp[i-1][j]+match[a[i]][' '],dp[i][j-1]+match[' '][b[j]]);
                dp[i][j]=max(dp[i-1][j-1]+match[a[i]][b[j]],dp[i][j]);
            }
        }
        printf("%d\n",dp[n][m]);
    }
    return 0;
}