DNA Sorting

DNA Sorting

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

刚开始思考的时候遇到一个问题。记录逆序数的数组和字符串数组无法一一对应，但是依靠乘以1000后加上原字符串位置可以完美这个问题。详见代码。

代码示例

#include<stdio.h>#include <string.h>#include <stdlib.h>int cmp(const void * a, const void * b) {    return((*(int*)a-*(int*)b));    }int main(){    char a[109][60];    int b[100]={0};    int i,j,c,n,m;    scanf("%d %d",&n,&m);//n is the lenght,m is the number    for(c=0;c<m;c++){        scanf("%s",&a[c]);        for(i=0;i<n;i++){            for(j=i;j<n;j++){                if(a[c][i]>a[c][j]){                    b[c]++;}            }        }        b[c]=b[c]*1000+c;        //printf("%d\n",b[c]);//以ASCII码比较大小    }    qsort(b,m,sizeof(b[0]),cmp);    for(i=0;i<m;i++){        printf("%s\n",a[b[i]%1000]);    }return 0;}

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