• Hdu 1080


    题目链接

    Human Gene Functions

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 2271    Accepted Submission(s): 1285


    Problem 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. 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

     Accepted Code:

     1 /*************************************************************************
     2     > File Name: D.cpp
     3     > Author: Stomach_ache
     4     > Mail: sudaweitong@gmail.com
     5     > Created Time: 2014年07月20日 星期日 20时04分25秒
     6     > Propose: 
     7  ************************************************************************/
     8 #include <map>
     9 #include <cmath>
    10 #include <string>
    11 #include <cstdio>
    12 #include <fstream>
    13 #include <cstring>
    14 #include <iostream>
    15 #include <algorithm>
    16 using namespace std;
    17 
    18 const int maxn = 105;
    19 char s[maxn], t[maxn];
    20 int dp[maxn][maxn];
    21 int a[5][5] = {{5, -1, -2, -1, -3}, {-1, 5, -3, -2, -4},
    22             {-2, -3, 5, -2, -2}, {-1, -2, -2, 5, -1},
    23               {-3, -4, -2, -1, 0}};
    24 
    25 
    26 int 
    27 max(int x, int y, int z) {
    28       return x > y ? 
    29                    x > z ? x : z
    30            : y > z ? y : z;
    31 }
    32 
    33 int
    34 main(void) {
    35 #ifndef ONLINE_JUDGE
    36       freopen("in.txt", "r", stdin);
    37 #endif
    38       map<char, int> match;
    39     match['A'] = 0;
    40     match['C'] = 1;
    41     match['G'] = 2;
    42     match['T'] = 3;
    43     match[' '] = 4;
    44       int c;
    45     scanf("%d", &c);
    46       while (c--) {
    47           int len1, len2;
    48         scanf("%d %s", &len1, s+1);
    49         scanf("%d %s", &len2, t+1);
    50         dp[0][0] = 0;
    51         for (int i = 1; i <=len1; i++) dp[i][0] = dp[i-1][0] + a[match[s[i]]][4];
    52         for (int i = 1; i <=len2; i++) dp[0][i] = dp[0][i-1] + a[match[t[i]]][4];
    53         for (int i = 1; i <= len1; i++) {
    54               for (int j = 1; j <= len2; j++) {
    55                   dp[i][j] = max(dp[i][j-1]+a[match[t[j]]][4], dp[i-1][j]+a[match[s[i]]][4], dp[i-1][j-1]+a[match[s[i]]][match[t[j]]]);
    56             }
    57         }
    58         printf("%d
    ", dp[len1][len2]);
    59     }
    60 
    61     return 0;
    62 }
  • 相关阅读:
    读取列表下标
    字典dict详解
    使用mysql的长连接
    oauth授权协议的原理
    安装性能测试工具:sysbench和使用apache的ab
    发送邮件出现问题
    获取用户的真实ip
    清理代码的阅读笔记
    开发中三个经典的原则
    要干大事就不能把面子看得太重
  • 原文地址:https://www.cnblogs.com/Stomach-ache/p/3857354.html
Copyright © 2020-2023  润新知