DNA Sorting

最新推荐文章于 2025-04-16 21:18:38 发布

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本文介绍了一种基于DNA字符串的排序性进行分类的方法。通过计算字符串中的逆序数来衡量其排序性，并据此对DNA字符串进行从最有序到最无序的排序。输入包括字符串长度和数量，输出为按排序性排列的DNA字符串列表。

Problem 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 Input10 6
AACATGAAGG
TTTTGGCCAA
TTTGGCCAAA
GATCAGATTT
CCCGGGGGGA
ATCGATGCAT
Sample OutputCCCGGGGGGA
AACATGAAGG
GATCAGATTT
ATCGATGCAT
TTTTGGCCAA
TTTGGCCAAA

#include<stdio.h>
int main()
{
    int n,m,i,j,k,s[200];
    char c[200][100],t;
    scanf("%d%d",&m,&n);
    for(i=0;i<n;i++)
    {
        s[i]=0;
        scanf("%s",c[i]);
        for(j=0;j<m;j++)//求逆序数
        {
            for(k=0;k<j;k++)
                if(c[i][j]<c[i][k])
                    s[i]++; 
        }
    } 
    for(i=0;i<n-1;i++)//对逆序数排序的同时要对DNA也排序
    {
        k=i;
        for(j=i+1;j<n;j++)  
            if(s[k]>s[j])
                k=j;
        if(k!=i)
        {
            for(j=0;j<m;j++)
            {
                t=c[k][j];
                c[k][j]=c[i][j];
                c[i][j]=t;
            }
            j=s[k];
            s[k]=s[i];
            s[i]=j;
        }
    }
    for(i=0;i<n;i++)
        printf("%s\n",c[i]);
 return 0;
}