POJ：DNA Sorting 特殊的排序

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

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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

很有意思的题目， POJ的题目都感觉相当好。

按照排序度的高低来排序，排序的排序？呵呵

因为数据的特殊性，所以本题可以计算逆序数的方法，可以0MS通过的，不过这里我使用的方法是：

1 merge sort来计算排序度

2 继续使用归并排序，排好最终序列

二次归并排序啊，呵呵，最终运行速度是16MS，做不到0MS，但是可以运行在无数据特殊性的情况下，通用性高。

#include <stdio.h>
#include <iostream>
#include <string>
#include <vector>
#include <algorithm>
#include <math.h>
using namespace std;  

struct DNA
{
    string s;
    int n;
    DNA(int i = 0, string ss = "") : n(i), s(ss){}
    bool operator<(const DNA &a) const
    {
        return n < a.n;
    }
};  

string biMerge(string &L, string &R, int &c)
{
    string s;
    unsigned i = 0, j = 0;
    while ( i < L.size() && j < R.size() )
    {
        if (L[i] <= R[j]) s.push_back(L[i++]);
        else
        {
            s.push_back(R[j++]);
            c += L.size() - i;
        }
    }
    if ( i < L.size() ) s.append(L.substr(i));
    else    s.append(R.substr(j));
    return s;
}  

void biSort(string &s, int &c)
{
    if (s.size() < 2) return ;
    int mid = ((s.size()-1)>>1);//错误：写成s.size()>>1  

    string L(s.substr(0, mid+1));
    biSort(L, c);  

    string R(s.substr(mid+1));
    biSort(R, c);  

    s = biMerge(L, R, c);
}  

void DNASorting()
{
    int Len = 0, T = 0, c = 0;
    cin>>Len>>T;
    vector<DNA> dnas(T);
    string t;
    for (int i = 0; i < T; i++)
    {
        cin>>t;
        dnas[i].s = t;
        c = 0;
        biSort(t, c);
        dnas[i].n = c;
    }
    sort(dnas.begin(), dnas.end());
    for (int i = 0; i < T; i++)
    {
        cout<<dnas[i].s<<endl;
    }
}  

int main()
{
    DNASorting();
    return 0;
}

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