HDU1379-逆序数排序 DNA Sorting

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概述

DNA Sorting

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1716 Accepted Submission(s): 823

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

Source

East Central North America 1998

思路：

题目意思是：根据DNA串的逆序数时行排序，如果DNA串的逆序数相等，则保持DNA原本的相对位置不变。

所以，重点是求出字符串的逆序数，求逆序数可以用蛮力法-O(n^2)，也可以用归并排序的方法-O(nlgn)。打开链接

#include <cstdio>
#include <string>
#include <iostream>
#include <algorithm>
using namespace std;

char arr[120],tmp[120];
int sum = 0;
struct node
{
	int id;
	int inversion;
	char str[120];
} dna[120];

bool cmp(const node& x,const node& y)
{
	if (x.inversion != y.inversion)
		return x.inversion < y.inversion;
	else
		return x.id < y.id;
}

void Merge(int left,int mid,int right)
{
	int i=left, j=mid+1, k=left;

	while(i <= mid && j <= right)
	{
		if(arr[i] <= arr[j])
			tmp[k++]=arr[i++];
		else
		{
			tmp[k++]=arr[j++];
			sum += mid-i+1;		//求逆序数
		}
	}
	while(i <= mid)
		tmp[k++]=arr[i++];
	while(j <= right)
		tmp[k++]=arr[j++];

	for(int i=left;i <= right;++i)
		arr[i]=tmp[i];
}	
void MergeSort(int left,int right)
{
	if(left < right)
	{
		int mid = (left + right) >> 1;
		MergeSort(left,mid);
		MergeSort(mid+1,right);
		Merge(left,mid,right);
	}
}

int main()
{
#ifndef ONLINE_JUDGE
	freopen("2.txt","r",stdin);
#endif
	int Case,n,m,i;
	scanf("%d ",&Case);
	while(Case--)
	{
		scanf("%d%d ",&n,&m);
		for (i=0; i < m; ++i)
		{
			scanf("%s ",arr);
			strcpy(dna[i].str,arr);
			sum = 0;
			MergeSort(0,n-1);
			dna[i].id = i;
			dna[i].inversion = sum;
		}
		sort(dna,dna+m,cmp);
		for(i=0; i < m; ++i)
			puts(dna[i].str);
	}
    return 0;
}