DNA Sorting(水题)

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

题解：首先求出逆序数，然后对逆序数进行排队，最后输出排序后的DNA序列。对于求逆序数，只需对序列进行遍历，找出本应出现在某字符之后，却出现在了它之前的字符数即可。

代码如下：

#include <stdio.h>

#include <stdlib.h>

#include <math.h>

#include <string.h>

int main()

{

    int n,m,i,j,k,s[150]= {0};

    char str[150][60],p[60];

    scanf("%d%d",&n,&m);

    for(i=0; i<m; i++)

    {

        scanf("%s",&str[i]);

        for(j=0; j<n-1; j++)

            for(k=j+1; k<n; k++)

                if(str[i][k]<str[i][j])

                    s[i]++;

    }

    for(i=0; i<m-1; i++)

        for(j=i+1; j<m; j++)

            if(s[i]>s[j])

            {

                strcpy(p,str[i]);

                strcpy(str[i],str[j]);

                strcpy(str[j],p);

                k=s[i];

                s[i]=s[j];

                s[j]=k;

                for(k=0; k<60; k++)

                    p[k]=0;

            }

    for(i=0; i<m; i++)

        printf("%s\n",str[i]);

    return 0;

}