北大ACM 1007题—DNA Sorting

DNA Sorting

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 73451		Accepted: 29334

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

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

Source

East Central North America 1998

#include<iostream>
#include<string>
#include <cstdlib> 
using namespace std;

class DNA{
public:
	string seqDNA;
	int num;
};

int compare(const void *p1, const void *p2)
{
	return(((DNA*)p1)->num-((DNA*)p2)->num);
}

int main()
{
	DNA *dna;
	int n,m,num,count(0);
	string listDNA;
	cin>>m>>n;
	dna=new DNA[n];

	while(count	<n)
	{
		cin>>listDNA;
		num=0;
		for(int i=0;i<m-1;++i)
		{                
			//寻找逆序数目
			for(int j=i+1;j<m;++j)
			{
				if(listDNA.at(i)>listDNA.at(j))
					++num;
			}
		}
		dna[count].num=num;
		dna[count++].seqDNA=listDNA;
	}

	//根据逆序数目进行快速排序
	qsort(dna,n,sizeof(DNA),compare);         
	for(int i=0;i<n;++i)
		cout<<dna[i].seqDNA<<endl;

	delete []dna;
	return 0;
}