1007. DNA Sorting

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

РЕШЕНИЕ

сортировка

Анализ алгоритма

Для каждой строки вычислим число инверсий в ней. Это число вместе с самой строкой храним в векторе пар. Выполняем стабильную сортировку строк в порядке неубывания количества инверсий в них и выводим результат.

Реализация алгоритма

#include <algorithm>

#include <cstdio>

#include <vector>

#include <string>

using namespace std;

int i, n, m, tests;

vector<pair<string,int> > s;

vector<pair<string,int> >::iterator iter;

char stroka[55];

int value(string s)

{

int i, j, res = 0;

for(i = 0; i < n - 1; i++)

for(j = i + 1; j < n; j++)

if (s[i] > s[j]) res++;

return res;

}

int lt(pair<string,int> x, pair<string,int> y)

{

return (x.second < y.second);

}

int main(void)

{

scanf("%d %d\n",&n,&m);

for(i = 0; i < m; i++)

{

gets(stroka);

s.push_back(make_pair(stroka,value(stroka)));

}

stable_sort(s.begin(),s.end(),lt);

for(iter = s.begin();iter != s.end();iter++)

puts((*iter).first.c_str());

return 0;

}