zoukankan      html  css  js  c++  java
  • DNA Sorting 分类: POJ 2015-06-23 20:24 9人阅读 评论(0) 收藏

    DNA Sorting
    Time Limit: 1000MS Memory Limit: 10000K
    Total Submissions: 88690 Accepted: 35644

    Description
    One measure of unsortedness'' in a sequence is the number of pairs of entries that are out of order with respect to each other. For instance, in the letter sequenceDAABEC”, this measure is 5, since D is greater than four letters to its right and E is greater than one letter to its right. This measure is called the number of inversions in the sequence. The sequence AACEDGG'' has only one inversion (E and D)---it is nearly sorted---while the sequenceZWQM” has 6 inversions (it is as unsorted as can be—exactly the reverse of sorted).

    You are responsible for cataloguing a sequence of DNA strings (sequences containing only the four letters A, C, G, and T). However, you want to catalog them, not in alphabetical order, but rather in order of sortedness'', frommost sorted” to “least sorted”. All the strings are of the same length.

    Input
    The first line contains two integers: a positive integer n (0 < n <= 50) giving the length of the strings; and a positive integer m (0 < m <= 100) giving the number of strings. These are followed by m lines, each containing a string of length n.

    Output
    Output the list of input strings, arranged from most sorted'' toleast sorted”. Since two strings can be equally sorted, then output them according to the orginal order.

    Sample Input

    10 6
    AACATGAAGG
    TTTTGGCCAA
    TTTGGCCAAA
    GATCAGATTT
    CCCGGGGGGA
    ATCGATGCAT

    Sample Output

    CCCGGGGGGA
    AACATGAAGG
    GATCAGATTT
    ATCGATGCAT
    TTTTGGCCAA
    TTTGGCCAAA
    大致的意思就是对DNA的逆序数排序

    #include <iostream>
    #include <string>
    #include <algorithm>
    #define RR freopen("input.txt","r",stdin)
    #define WW freopen("ouput.txt","w",stdout)
    
    using namespace std;
    const int INF=0x3f3f3f3f;
    int n,m;
    struct DNA
    {
        string str;
        int num ;
        void NUM()
        {
            num=0;
            for(int i=0;i<n;i++)
            {
                int sum=0;
                for(int j=i-1;j>=0;j--)
                {
                    if(str[i]<str[j])
                    {
                        sum++;
                    }
                }
                num+=sum;
            }
        }
    }D[110];
    bool cmp(DNA a,DNA b)
    {
        return a.num<b.num;
    }
    int main()
    {
       cin>>n>>m;
       for(int i=0;i<m;i++)
       {
           cin>>D[i].str;
           D[i].NUM();
       }
       sort(D,D+m,cmp);
       for(int i=0;i<m;i++)
       {
           cout<<D[i].str<<endl;
       }
        return 0;
    }
    

    版权声明:本文为博主原创文章,未经博主允许不得转载。

  • 相关阅读:
    linux系统中rsync+inotify实现服务器之间文件实时同步
    用Nginx搭建CDN服务器方法-开启Nginx缓存与镜像,自建图片服务器
    CentOS 搭建dns服务器 解析任意域名
    批量取控件的值
    我的一类库
    asp.net相关的一些代码
    C#的一些代码
    口算训练(唯一分解定理 + 二分+2018年女生赛)
    Codeforces Round #484 (Div. 2)
    Codeforces Round #483 (Div. 2) [Thanks, Botan Investments and Victor Shaburov!]
  • 原文地址:https://www.cnblogs.com/juechen/p/4721997.html
Copyright © 2011-2022 走看看