POJ【DNA Sorting】

Phr_Nick

于 2019-01-22 10:18:06 发布

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分类专栏：算法，刷题文章标签： POJ STL

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本文介绍了一种基于冒泡排序思想的逆序数计算方法，用于衡量DNA序列的有序程度，并通过C++ STL中的multimap实现了从最有序到最无序的DNA序列排序。

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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 sequence ``DAABEC'', 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 sequence ``ZWQM'' 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'', from ``most 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'' to ``least 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

思路：1、将数据存入multimap<string,int> 的键值对中。

2、利用冒泡排序的思想，来求逆序的数量。

3、根据逆序的数量，对multimap<string,int> 的键值对进行修改。

4、将multimap<string,int>的数据，存入multimap<int,string>中，这样可以偷个懒，不需要自己实现排序输出。

【注：】这个算法不是最优的，我是为了自己复习STL的相关操作。爱你哟，笔芯，欢迎交流。

#include <iostream>
#include <string>
#include <map>
#include <algorithm>
using namespace std;
int countFunction(string data)//利用冒泡排序的思路来计算逆序数
{
    int Count=0;
    for(int i=0;i<data.length();i++)
    {
        for(int j=i+1;j<data.length();j++){
            if(data[i]>data[j])
                Count++;
        }
    }
    return Count;
}
int main()
{
    /*
     * 输入相关数据
    */
    int m,n;
    cin>>m>>n;
    multimap<string,int> data;
    multimap<int,string>result;
    while (n--) {
        string temp;
        cin>>temp;
        data.insert(pair<string,int>(temp,0));
    }

    /*
     * 更新将每个字符串对应的逆序数
    */
    multimap<string,int>::iterator iter;
    for(iter=data.begin();iter!=data.end();iter++)
    {
        iter->second=countFunction(iter->first);
    }


    /*
     * multimap<string,int> 转存到multimap<int,string>;利用关联容器会对键所在键值对进行排序。
    */
    for(iter=data.begin();iter!=data.end();iter++)
    {
        result.insert(pair<int,string>(iter->second,iter->first));
    }

    /*
     * 直接输出
    */
    multimap<int,string>::iterator iter2;
    for(iter2=result.begin();iter2!=result.end();iter2++)
    {
        cout<<iter2->second<<endl;
    }
    return 0;
}