HDU 1379 DNA sorting(求逆序数)

　　　　　　　　　　　　　　　DNA Sorting

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

 

虽然此题傻暴力求逆序数即可，但是归并排序更快。

 

代码：

#include<cstdio>
#include<cstring>
#include<malloc.h>
#include<algorithm>
using namespace std;

int ans;

struct str
{
    char s[105];
    char sort_s[105];
    int Num;
};

bool cmp(struct str a,struct str b)
{
    return a.Num<b.Num;
}

void MergeArray(char a[],int L,int Mid,int R,char temp[])
{
    int k=0;
    int i=L,j=Mid+1;
    int m=Mid,n=R;
    while(i<=m&&j<=n)
    {
        if(a[i]<=a[j])
        temp[k++]=a[i++];
        else
        {
      　　  temp[k++]=a[j++];
      　　  ans+=m-i+1;
        }
    }
    while(i<=m)
    temp[k++]=a[i++];
    while(j<=n)
    temp[k++]=a[j++];
    for(i=0;i<k;i++)
    a[L+i]=temp[i];
}

void MergeSort(char a[],int L,int R,char temp[])
{
    if(L<R)
    {
        int Mid=(L+R)/2;
        MergeSort(a,L,Mid,temp);
        MergeSort(a,Mid+1,R,temp);
        MergeArray(a,L,Mid,R,temp);
    }
}

int main()
{
    struct str S[55];
    int i,m,n,t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&m,&n);
        for(i=0;i<n;i++)
        scanf("%s",S[i].s);
        for(i=0;i<n;i++)    
        {
            ans=0;
            int len=strlen(S[i].s);
            strcpy(S[i].sort_s,S[i].s);
            char *p=(char*)malloc(sizeof(char)*(len+1));
            MergeSort(S[i].sort_s,0,len-1,p);
            S[i].Num=ans;
            free(p);
        }
        sort(S,S+n,cmp);
        for(i=0;i<n;i++)
        printf("%s
",S[i].s);
    }
    return 0;
}