DNA Sorting

DNA Sorting

时间: 1ms        内存:64M

描述:

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.

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.

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.

示例输入:

10 6
AACATGAAGG
TTTTGGCCAA
TTTGGCCAAA
GATCAGATTT
CCCGGGGGGA
ATCGATGCAT

示例输出:

CCCGGGGGGA
AACATGAAGG
GATCAGATTT
ATCGATGCAT
TTTTGGCCAA
TTTGGCCAAA

提示:

参考答案(内存最优[752]):

#include<stdio.h>
#include<string.h>
int main()
{
    int n,m,i,j,s[105]={0},q,max=0;
    char a[200][200];
    scanf("%d%d",&n,&m);
    getchar();
    for(i=0; i<m; i++)
        gets(a[i]);
    for(i=0; i<m; i++)
    {
        for(j=0; j<n; j++)
            for(q=j+1; q<n; q++)
                if(a[i][j]>a[i][q])
                    s[i]++;
        if(max<s[i])
            max=s[i];
    }
    for(i=0; i<=max; i++)
        for(j=0; j<m; j++)
            if(s[j]==i)
                puts(a[j]);
    return 0;
}

参考答案(时间最优[0]):

#include <stdio.h>
#include <algorithm>
#include <iostream>
#include<string.h>
#include<math.h>
using namespace std;
struct DNA
{
    char c[55];
    int data;
} d[105];
int nixu(char *c)
{
    int k=0;
    int n=strlen(c);
    for(int i=0; i<n; i++)
    {
        for(int j=i+1; j<n; j++)
            if(c[i]>c[j])k++;
    }
    return k;
}
void paixu(int m)
{
    for(int i=0; i<m; i++)
        for(int j=0; j<m-i-1; j++)
            if(d[j+1].data<d[j].data)
            {
                DNA a=d[j+1];
                d[j+1]=d[j];
                d[j]=a;
            }
}
int main()
{
    int n,m;
    cin>>n>>m;
    getchar();
    for(int i=0; i<m; i++)
    {
        scanf("%s",d[i].c);
        d[i].data=nixu(d[i].c);
    }
    paixu(m);
    for(int i=0; i<m; i++)
        puts(d[i].c);
    return 0;
}

题目和答案均来自于互联网,仅供参考,如有问题请联系管理员修改或删除。

点赞

发表评论

电子邮件地址不会被公开。必填项已用 * 标注