POJ 1080_Human Gene Functions
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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.
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
1421
题意:两组字符序列进行配对,字符配对对应的值为表中所示,每组字符的前后顺序不能改动,相邻字符间可以插入任意多“-”,求一种情况使得配对值的和最大。
解题思路:DP的方法,我们可以发现求解的过程是可以分阶段的,并且满足每一阶段的决策都是独立的,基于上一阶段的最佳决策。举例,序列s1 s2最后的解是通过两个序列的子序列s1-1 s2-1d的最佳解和s1[end] s2[end]配对值组成,那么s1[end] s2[end]有可能是字母,有可能是‘-’,也就是三种情况:字母配‘-’,字母配字母,‘-’配字母。整理成DP的公式就是:dp[i][j]= max(dp[i-1][j-1]+val[s1[end]][s2[end]], dp[i][j-1]+val['-'][s2[end], dp[i-1][j]+val[s1[end]]['-']]);
有了动态规划的流程,有了公式,我们就可以开始DP了。
#include <iostream>using namespace std;#define MAX 105int dp[MAX][MAX];int matrix[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 getID(char a){switch (a){case 'A': return 0;case 'C': return 1;case 'G': return 2;case 'T': return 3;case '-': return 4;}}int max(int a,int b,int c){ if(a>=b && a>=c) return a; if(b>=a && b>=c) return b; return c; } int main(){char s1[MAX];char s2[MAX];int times, len1, len2;//输入次数cin>>times;while (times--){//输入数据cin>>len1>>s1>>len2>>s2;//初始化决策表memset(dp,0,sizeof(dp)); dp[0][0] = 0;//计算i,j分别为0的dpfor (int i=1; i<=len1; i++)dp[i][0] = dp[i-1][0] + matrix[getID(s1[i-1])][4];for (int i=1; i<=len2; i++)dp[0][i] = dp[0][i-1] + matrix[4][getID(s2[i-1])];//DP,每一阶段的决策中都有3种情况,取其中最好的一种for (int i=1; i<=len1; i++)for (int j=1; j<=len2; j++){int tmp1 = dp[i-1][j] + matrix[getID(s1[i-1])][4];int tmp2 = dp[i][j-1] + matrix[4][getID(s2[j-1])];int tmp3 = dp[i-1][j-1] + matrix[getID(s1[i-1])][getID(s2[j-1])];dp[i][j] = max(tmp1, tmp2, tmp3);}cout<<dp[len1][len2]<<endl;}return 0;}
- POJ 1080_Human Gene Functions
- POJ 1080 Human Gene Functions
- POJ 1080 Human Gene Functions
- POJ 1080 Human Gene Functions
- poj 1080 Human Gene Functions
- POJ 1080 Human Gene Functions
- Poj 1080 Human Gene Functions
- poj 1080 Human Gene Functions
- poj 1080 Human Gene Functions
- POJ 1080 Human Gene Functions
- poj 1080 human gene functions
- POJ 1080 Human Gene Functions
- POJ 1080 Human Gene Functions
- POJ-1080-Human Gene Functions
- POJ 1080 Human Gene Functions
- POJ 1080 Human Gene Functions
- poj 1080 Human Gene Functions
- poj 1080 Human Gene Functions
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