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 该题 利用数组的技巧 巧算逆序数 如函数f所示运用结构体将 逆序数 与 字符串联系起来运用algorithm中的stable_sort函数 排序 相同时，不改变原有序列！！
#include<iostream>
#include<algorithm>
using namespace std;
struct DNA{
char str[55];
int num;
}d[105];
bool cmp(DNA a,DNA b)
{
return a.num<b.num;
}
int f(char s[],int n)
{
int a[5]={0,0,0,0},m=0;
for(int i=n-1;i>=0;i--){
switch(s[i]){
case ‘A‘:
a[1]++;
a[2]++;
a[3]++;
break;
case ‘C‘:
a[2]++;
a[3]++;
m+=a[1];
break;
case ‘G‘:
a[3]++;
m+=a[2];
break;
case ‘T‘:
m+=a[3];
break;
}
}
return m;
}
int main() {
int n,m;
cin>>n>>m;
for(int i=0;i<m;i++){
for(int j=0;j<n;j++)cin>>d[i].str[j];
d[i].num=f(d[i].str,n);
}
stable_sort(d,d+m,cmp);
for(int i=0;i<m;i++) {
for(int j=0;j<n;j++)cout<<d[i].str[j];
cout<<endl;
}
//system("pause");
return 0;
}