3915 - new DNA sorting

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. 

题目输入

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.

题目输出

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.

输入/输出样例

题目输入

10 6
AACATGAAGG
TTTTGGCCAA
TTTGGCCAAA
GATCAGATTT
CCCGGGGGGA
ATCGATGCAT

题目输出

CCCGGGGGGA
AACATGAAGG
GATCAGATTT
ATCGATGCAT
TTTTGGCCAA
TTTGGCCAAA

C语言解答

#include <stdio.h>
typedef struct DNAMAP
{
    char map[100];
    int sum;
}DNA;
int main()
{
    int m,n,i,j,k;
    DNA sort[50],t;
    scanf("%d%d\n",&m,&n);
    i=0;
    for (i=0;i<n;i++)
    {
        scanf("%s",sort[i].map);
        sort[i].sum=0;
        for (j=0;j<m-1;j++)
            for (k=j+1;k<m;k++)
            if (sort[i].map[j]>sort[i].map[k])
                sort[i].sum++;
    }
   for (i=0;i<n-1;i++)
        for (j=i+1;j<n;j++)
        if (sort[i].sum>=sort[j].sum)
        {
            t=sort[i];
            sort[i]=sort[j];
            sort[j]=t;
        }
    for (i=0;i<n-1;i++)
        printf("%s\n",sort[i].map);
    printf("%s",sort[n-1].map);
    return 0;
}

C++解答

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

typedef struct
{
    string dna;
    int count;
}DNA;
DNA dna[101];
int cmp(const void *a,const void *b)
{
    DNA *aa = (DNA *)a;
    DNA *bb = (DNA *)b;
    return aa->count-bb->count;
}
int main()
{
    int n,m;
    char c;
    cin>>n>>m;
    for(int i = 0; i < m; i++)
    {
        cin>>dna[i].dna;
        dna[i].count = 0;
        for(int j = 0; j < n; j++)
        for(int k = j+1; k < n; k++)
        {
            if(dna[i].dna[j]>dna[i].dna[k])
            dna[i].count++;
        }
    }
    qsort(dna,m,sizeof(dna[0]),cmp);
    for(int i = 0; i < m; i++)
    cout<<dna[i].dna<<endl;
    return 0;
}
时间限制 1 秒
内存限制 128 MB
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