[DP]HOJ 1316Human Gene Functions
2014-01-27 17:29
393 查看
传送门:Human Gene Functions
Submitted : 418, Accepted : 276
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. 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
Sample Output
解题报告:
此题属于LCS类的问题,但是状态转移方程不是我推出来的。
状态转移方程如下:
1.c[i][j]=max(c[i-1][j-1]+value(a[i],b[j]),c[i][j-1]+value('-',b[j]),c[i-1][j]+value(a[i],'-')) (i,j>0)
2.c[i][0] = c[i-1][0] + value[i]['-']
3.c[0][i] = c[0][i-1] + value['-'][i]
代码如下:
#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
int Achar[1005]; //A串
int Bchar[1005]; //B串
char temp[1005];
int dp[1005][1005];
int value[6][6] = {0,0,0,0,0,0,
0,5,-1,-2,-1,-3,
0,-1,5,-3,-2,-4,
0,-2,-3,5,-2,-2,
0,-1,-2,-2,5,-1,
0,-3,-4,-2,-1,0 };
int main(){
int t;
int At,Bt;
scanf("%d",&t);
while(t--){
scanf("%d%s",&At,temp);
for(int i=0;i<At;i++){
if(temp[i]=='A')
Achar[i]=1;
else if(temp[i]=='C')
Achar[i]=2;
else if(temp[i]=='G')
Achar[i]=3;
else if(temp[i]=='T')
Achar[i]=4;
}
scanf("%d%s",&Bt,temp);
for(int i=0;i<Bt;i++){
if(temp[i]=='A')
Bchar[i]=1;
else if(temp[i]=='C')
Bchar[i]=2;
else if(temp[i]=='G')
Bchar[i]=3;
else if(temp[i]=='T')
Bchar[i]=4;
}
memset(dp,0,sizeof(dp));
for(int i=1;i<Bt;i++)
dp[0][i]=dp[0][i-1]+value[5][Bchar[i-1]];
for(int i=1;i<At;i++)
dp[i][0]=dp[i-1][0]+value[Achar[i-1]][5];
for(int i=1;i<=At;i++)
for(int j=1;j<=Bt;j++)
dp[i][j]=max(dp[i-1][j-1]+value[Achar[i-1]][Bchar[j-1]],max(dp[i][j-1]+value[5][Bchar[j-1]],dp[i-1][j]+value[Achar[i-1]][5]));
printf("%d\n",dp[At][Bt]);
}
return 0;
}
Human Gene Functions
My Tags | (Edit) |
---|
Source : ACM ICPC Taejon(S.Korea) Regional Contest 2001 | |||
Time limit : 1 sec | Memory limit : 32 M |
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. 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
解题报告:
此题属于LCS类的问题,但是状态转移方程不是我推出来的。
状态转移方程如下:
1.c[i][j]=max(c[i-1][j-1]+value(a[i],b[j]),c[i][j-1]+value('-',b[j]),c[i-1][j]+value(a[i],'-')) (i,j>0)
2.c[i][0] = c[i-1][0] + value[i]['-']
3.c[0][i] = c[0][i-1] + value['-'][i]
代码如下:
#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
int Achar[1005]; //A串
int Bchar[1005]; //B串
char temp[1005];
int dp[1005][1005];
int value[6][6] = {0,0,0,0,0,0,
0,5,-1,-2,-1,-3,
0,-1,5,-3,-2,-4,
0,-2,-3,5,-2,-2,
0,-1,-2,-2,5,-1,
0,-3,-4,-2,-1,0 };
int main(){
int t;
int At,Bt;
scanf("%d",&t);
while(t--){
scanf("%d%s",&At,temp);
for(int i=0;i<At;i++){
if(temp[i]=='A')
Achar[i]=1;
else if(temp[i]=='C')
Achar[i]=2;
else if(temp[i]=='G')
Achar[i]=3;
else if(temp[i]=='T')
Achar[i]=4;
}
scanf("%d%s",&Bt,temp);
for(int i=0;i<Bt;i++){
if(temp[i]=='A')
Bchar[i]=1;
else if(temp[i]=='C')
Bchar[i]=2;
else if(temp[i]=='G')
Bchar[i]=3;
else if(temp[i]=='T')
Bchar[i]=4;
}
memset(dp,0,sizeof(dp));
for(int i=1;i<Bt;i++)
dp[0][i]=dp[0][i-1]+value[5][Bchar[i-1]];
for(int i=1;i<At;i++)
dp[i][0]=dp[i-1][0]+value[Achar[i-1]][5];
for(int i=1;i<=At;i++)
for(int j=1;j<=Bt;j++)
dp[i][j]=max(dp[i-1][j-1]+value[Achar[i-1]][Bchar[j-1]],max(dp[i][j-1]+value[5][Bchar[j-1]],dp[i-1][j]+value[Achar[i-1]][5]));
printf("%d\n",dp[At][Bt]);
}
return 0;
}
相关文章推荐
- 基于Android中dp和px之间进行转换的实现代码
- Android中dip、dp、sp、pt和px的区别详解
- C++ 动态规划
- DP(动态规划) 解游轮费用问题
- 关于爬楼梯的动态规划算法
- 动态规划 --- hdu 1003 **
- Android根据分辨率进行单位转换-(dp,sp转像素px)
- android 尺寸 dp,sp,px,dip,pt详解
- 字符串编辑距离
- ACM常用算法
- LightOJ 1013 - Love Calculator(DP)
- 动态规划求解最大连续子序列和
- TopCoder SRM 569 DIV2 Level3: MegaFactorialDiv2
- Topcoder SRM 547 DIV1 Level 1: Pillars
- TopCoder SRM DIV2 Level 3: RelativelyPrimeSubset
- HDU4758 AC自动机+DP (HDU4758与HDU2222)
- [LeetCode] Distinct Subsequences
- [LeetCode] Interleaving String
- [LeetCode] Unique Binary Search Tree
- [LeetCode] Edit Distance