POJ1007,DNA Sorting，排序水题

最新推荐文章于 2021-05-19 21:53:58 发布

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本文深入探讨了排序算法的应用场景及实现方式，通过实例展示了如何根据特定规则对DNA序列进行排序，强调了算法效率和复杂度的重要性。文章还提供了代码实现，详细解释了每一部分的功能，旨在帮助读者理解并掌握排序算法在实际问题解决中的应用。

DNA Sorting

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.

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 (0 < 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''. Since two strings can be equally sorted, then output them according to the orginal order.
Sample Input

10 6
AACATGAAGG
TTTTGGCCAA
TTTGGCCAAA
GATCAGATTT
CCCGGGGGGA
ATCGATGCAT

Sample Output

CCCGGGGGGA
AACATGAAGG
GATCAGATTT
ATCGATGCAT
TTTTGGCCAA
TTTGGCCAAA

分析：

一道很水的排序题，主要是按题目给出的乱序规则给每个串增加一个乱序属性值。

用Inv函数算出，之后排序就可以了。

code:

#include<iostream>
#include<cstdio>
#define MAX_LEN 50
#define MAX_NUM 100
using namespace std;
char DNA[MAX_NUM+5][MAX_LEN+5];
int inversion[MAX_NUM+5];
int order[MAX_NUM+5];
int Inv(char *s,int len)
{
    int count=0,i,j;
    for(i=0;i<len;i++)
        for(j=i+1;j<len;j++)
        {
            if(s[i]>s[j]) count++;
        }
    return count;
}
int main()
{
    int len,n,i,j;
    cin>>len>>n;
	getchar();
    for(i=0;i<n;i++)
    {
        gets(DNA[i]);
        inversion[i+1]=Inv(DNA[i],len);
    }
    inversion[0]=10000;
    order[0]=0;
    for(i=1;i<=n;i++)
    {
        order[i]=i;
        for(j=1;j<=n;j++)
            if(inversion[j]>inversion[order[i]]&&inversion[order[i]]<inversion[order[i-1]])
                order[i]=j;
    }
    for(i=0;i<n;i++)
        puts(DNA[order[i+1]]);
    return 0;
}