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.
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
题意大概是：设一个字母序列”DAABEC“的逆序数是5，由于D比它右边的4个字母大，而E比它右边的1个字母大。序列”AACEDGG“的逆序数是1。差点儿已经排好序
如今对DNA字符串序列进行分类。然而，分类不是按字母顺序，而是按”排序“的次序，从最多已知排序到最少已知排序
#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
struct aa
{
char s[55];
int x;
};
bool cmp(aa m,aa n)
{
return m.x<n.x;
}
int main()
{
int m,n,i,j,k;
char s[10];
cin>>n>>m;
aa a[110];
gets(s);
for(i=0; i<m; ++i)
{
gets(a[i].s);
a[i].x=0;
for(j=0; j<n; ++j)
for(k=j+1; k<n; k++)
{
if(a[i].s[j]>a[i].s[k])
a[i].x++;
}
}
sort(a,a+m,cmp);
for(i=0;i<m;++i)
cout<<a[i].s<<endl;
return 0;
}