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.
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
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
#include
#include
#include
#include
#include
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 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<