poj1007 -- DNA Sorting

最新推荐文章于 2021-05-19 21:53:58 发布

BestFSQ

最新推荐文章于 2021-05-19 21:53:58 发布

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本文介绍了一种用于评估DNA字符串“排序性”的算法，通过计算序列中逆序对的数量来衡量其排序程度，并给出了一段C++实现代码示例。

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DNA Sorting

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 100566		Accepted: 40371

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

思路就是创建一个结构体保存数据，求出每一个DNA的measure，然后根据measure从小到大排序，最后输出。

代码如下：

#include <algorithm>
#include <iostream>
#include <cstdio>
#include <string>
#include <vector>
using namespace std;

struct Node
{
    string DNA;
    int measure;

    Node(string str,int mea):DNA(str),measure(mea){}
};

bool cmp(Node node1,Node node2)  //按measure从小到大排序
{
    return node1.measure<node2.measure;
}

vector<Node> v;

int main()
{
    //freopen("test.txt","r",stdin);
    v.clear();
    int n,m;
    scanf("%d%d",&n,&m);
    while(m--)
    {
        string str;
        cin>>str;
        int measure=0;
        int len=str.length();
        for(int i=0;i<len-1;i++)
        {
            for(int j=i+1;j<len;j++)
            {
                if(str[i]>str[j])
                    measure++;
            }
        }
        Node node(str,measure);
        v.push_back(node);
    }
    sort(v.begin(),v.end(),cmp);
    for(int i=0;i<v.size();i++)
        cout<<v[i].DNA<<endl;
    return 0;
}