hdu1379 DNA Sorting

链接：http://acm.hdu.edu.cn/showproblem.php?pid=1379

DNA Sorting

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4034    Accepted Submission(s): 1765

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

1 10 6 AACATGAAGG TTTTGGCCAA TTTGGCCAAA GATCAGATTT CCCGGGGGGA ATCGATGCAT

 

Sample Output

CCCGGGGGGA AACATGAAGG GATCAGATTT ATCGATGCAT TTTTGGCCAA TTTGGCCAAA    

#include<stdio.h>
#define N 200
int main()
{
	int len,n,i,j,h;
	struct dna{
		char str[N];
		int number;
	}a[N],t;
	scanf("%d%d",&len,&n);
	for(i=0;i<n;i++)
	{
		scanf("%s",a[i].str);
		a[i].number=0;
	}
	
	for(i=0;i<n;i++)
	{
		for(j=0;j<len;j++)
		{
			for(h=j;h<len;h++)
			if(a[i].str[h]<a[i].str[j])
			a[i].number++;
		}
	} 
	for(i=0;i<n;i++)
	{
		for(j=0;j<n-1;j++)
		if(a[j].number>a[j+1].number)
		{
			t=a[j];
			a[j]=a[j+1];
			a[j+1]=t;
		}
	}
	for(i=0;i<n;i++)
	printf("%s
",a[i].str);
	return 0;
} 

主要意思就是判断逆序数，从小到大排序，用结构体可以解决，另外，逆序数有可能相同，需要用稳定排序