HDU1379:DNA Sorting

Problem 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

110 6AACATGAAGGTTTTGGCCAATTTGGCCAAAGATCAGATTTCCCGGGGGGAATCGATGCAT

 

Sample Output

CCCGGGGGGAAACATGAAGGGATCAGATTTATCGATGCATTTTTGGCCAATTTGGCCAAA

 

 

 

一直没弄懂题意，看了人家的解题报告才知道，原来就是看每个字母后面有几个字母比它大，然后再把数目全加起来，按总和的升序排列

#include<stdlib.h>#include<stdio.h>#include <string>#include <iostream>#include <algorithm>#include <cstdio>using namespace std;struct DNA{    string str;    int count;} w[1001];bool comp(DNA x,DNA y){    return x.count<y.count;}int main(){    int s,n,i,j,k,t;    cin >> t;    while(t--)    {        scanf("%d %d",&s,&n);        for(i=0; i<n; i++)        {            cin>>w[i].str            w[i].count=0;            for(j=0; j<=s-2; j++) //**选择排序**//            {                for(k=j+1; k<=s-1; k++)                {                    if(w[i].str[j]>w[i].str[k]) w[i].count++;                }            }        }        stable_sort(w,w+n,comp);        for(i=0; i<n; i++)        {            cout<<w[i].str<<endl;        }    }    return 0;}