E

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.

This problem contains multiple test cases!

The first line of a multiple input is an integer N, then a blank line followed by N input blocks. Each input block is in the format indicated in the problem description. There is a blank line between input blocks.

The output format consists of N output blocks. There is a blank line between output blocks.

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 (1 < 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''. If two or more strings are equally sorted, list them in the same order they are in the input file.

Sample Input

1

10 6
AACATGAAGG
TTTTGGCCAA
TTTGGCCAAA
GATCAGATTT
CCCGGGGGGA
ATCGATGCAT

Sample Output

CCCGGGGGGA
AACATGAAGG
GATCAGATTT
ATCGATGCAT
TTTTGGCCAA
TTTGGCCAAA

题意：

就是 找出每个字母往后比他小的个数然后求和，从小到大排列

分析：

就是一个一个的往后找

代码：

#include<bits/stdc++.h>
using namespace std;
int Vp(string a,int n)
{
    int i,j,s=0;
    for(i=0;i<n;i++)
        for(j=i+1;j<n;j++){
            if(a[i]>a[j])
                s++;
        }
    return s;
}
int main()
{
    string a[105];
    int n,m,t,i,p,s=0,r;
    multimap<int,int>v;
    multimap<int,int>::iterator it;
    while(cin>>t)
    {
        s=0;
        r=0;
        while(cin>>n>>m&&r<t)
        {
            if(s)
                cout<<endl;
            else
                s=1;
            r++;
            for(i=0;i<m;i++){
                cin>>a[i];
                p=Vp(a[i],n);
                v.insert(make_pair(p,i));
            }
            for(it=v.begin();it!=v.end();it++)
                cout<<a[(*it).second]<<endl;
            v.clear();
        }
    }


}

感受：

就是英文不好看。。。英语内伤啊