PKU ACM 1080-human gene function

题目来源：http://acm.pku.edu.cn/JudgeOnline/problem?id=1080
解题报告：最近模糊的看了一下算法导论的动态规划一章，好像理解了，结果做题的时候，发现还是不会做，总是找不到子问题，不知道怎么设最优子结构，本来想找一道简单的来做，就找到了1080，结果还是没思路，看了别人的代码，才理解了一点，看来还是要多做多练啊。
这道题是求两个字符串的相似度，设第一个字符串s1[0],s1[1],s1[2],....s1[len1-1], 第二个字符串s2[0],s2[1],....s2[len2-1]，那么看两个字符串的最后一个字符的关系，有三种可能性：
1. s1[len1-1]与s2[len2-1]对应
2. s1[len1-1]与‘-’对应
3. s2[len2-1]与'-'对应
对第一种情况，字符串的相似度就是s1[0],s1[1],s1[2],....s1[len1-2] 与 s2[0],s2[1],....s2[len2-2]的相似度+s1[len1-1]与s2[len2-1]间的相似度。显然，要达到最大值，必须使s1[0],s1[1],s1[2],....s1[len1-2] 与 s2[0],s2[1],....s2[len2-2]这两个字符串的相似度也达到最大值。
对第二种情况，两个字符串的相似度就是s1[0],s1[1],s1[2],....s1[len1-2] 与 s2[0],s2[1],....s2[len2-1]的相似度+s1[len1-1]与‘-’间的相似度。显然，要达到最大值，必须使s1[0],s1[1],s1[2],....s1[len1-2] 与 s2[0],s2[1],....s2[len2-1]这两个字符串的相似度也达到最大值。
对第三种情况，两个字符串的相似度就是s1[0],s1[1],s1[2],....s1[len1-1] 与 s2[0],s2[1],....s2[len2-2]的相似度+'-'与s2[len2-1]间的相似度。显然，要达到最大值，必须使s1[0],s1[1],s1[2],....s1[len1-1] 与 s2[0],s2[1],....s2[len2-2]这两个字符串的相似度也达到最大值。
利用动态规划自底向上，设f[i][j]代表字符串s1的前i个字符所组成的字符串与字符串s2的前j个字符所组成的字符串间的相似度，那么
f[i][j]=max(
f[i-1][j-1]+value(s1[i-1],s2[j-1]), //第一种情况
f[i-1][j] + value(s1[i-1],'-'), //第二种情况
f[i][j-1] + value('-', s2[j-1]) //第三种情况
)

#include <iostream>
using namespace std;

#define INF 100

int f[102][102];

int v[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,-INF
};

int toInt(char c)
{
	int i;
	switch(c)
	{
	case 'A': i=0;break;
	case 'C': i=1;break;
	case 'G': i=2;break;
	case 'T': i=3;break;
	case '-': i=4;break;
	default: break;
	}
	return i;
}

int value(char a, char b)
{
	int m=toInt(a);
	int n=toInt(b);
	return v[m]
;
}

int max(int a,int b, int c)
{
	int m=a<b? b: a;
	m=m<c? c: m;
	return m;
}

int main()
{
	int caseNum;
	cin >> caseNum;
	while(caseNum--)
	{
		int len1,len2;
		cin >> len1;
		char *s1=new char[len1+1];
		cin >> s1;
		cin >> len2;
		char *s2=new char[len2+1];
		cin >> s2;
		for(int i=0;i<=len1;i++)
		{
			for(int j=0;j<=len2;j++)
			{
				if(i==0 && j==0)
					f[i][j]=0;
				else if(i==0)
					f[i][j]=f[i][j-1]+value('-',s2[j-1]);
				else if(j==0)
					f[i][j]=f[i-1][j]+value(s1[i-1],'-');
				else
				{
					f[i][j]=max(f[i-1][j-1]+value(s1[i-1],s2[j-1]), f[i-1][j]+value(s1[i-1],'-'), f[i][j-1]+value('-',s2[j-1]));
				}
			}
		}
		cout <<  f[len1][len2] << endl;
	}
}

附录：
Human Gene Functions

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 9149		Accepted: 5045

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.
一张对应的表格，图片显示不出来。。
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