Problem 62: DNA Sorting（简单逆序数， 排序问题）

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. InputThe 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.OutputOutput 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 InputSample Input10 6AACATGAAGGTTTTGGCCAATTTGGCCAAAGATCAGATTTCCCGGGGGGAATCGATGCATSample OutputCCCGGGGGGAAACATGAAGGGATCAGATTTATCGATGCATTTTTGGCCAATTTGGCCAAA

#include<iostream>#include <stdio.h>using namespace std;typedef struct {    char square[100];    int n;} Sam;Sam a[100];int main(void) {    int i, j, k, m, n;    scanf("%d%d", &m, &n);    getchar();    for(i=0; i<n; i++) {        for(j=0; j<m; j++)            scanf("%c",&a[i].square[j]);        a[i].n = 0;        getchar();        for(j=0; j<m-1; j++) {            for(k=j+1; k<m; k++) {                if(a[i].square[j] > a[i].square[k])                    a[i].n += 1;            }        }    }    for(i=0; i<n-1; i++) {        for(j=0; j<n-i-1; j++) {            if(a[j].n > a[j+1].n) {                Sam temp = a[j];                a[j] = a[j+1];                a[j+1] = temp;            }        }    }    for(i=0; i<n; i++) {        for(j=0; j<m; j++)            printf("%c", a[i].square[j]);        puts("");    }    return 0;}