DNA Sorting（OpenJ_Bailian - 1007）

最新推荐文章于 2019-10-18 23:40:42 发布

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最新推荐文章于 2019-10-18 23:40:42 发布

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分类专栏：脑力思维小组、个人赛题目文章标签：快排字符串

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博客围绕DNA字符串排序问题展开，介绍了衡量序列“无序度”的逆序数概念。要求对DNA字符串按“有序度”从高到低排序，给出了输入输出格式及示例，还提示解题思路是统计每排字母逆序数并排序，最后给出代码。

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B - DNA Sorting

来源： OpenJ_Bailian - 1007

Time limit：1000 ms Memory limit：65536 kB

OS：Linux Source：East Central North America 1998

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.

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

Hint：题目要求就是找出每一排中，每一个字母后边比它大的（ASC码或者字母序）字母的数目，然后每一排中所有字母之和。然后将每一排按从小到大排序；题目并不难，开始以为会超时，应该是数据量不是很大。

代码如下：

#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
using namespace std;
char s[110][55];
int a[110] = {0};
void qs(int l, int r);
int main()
{
    int n, m, x, i, j, k;
    scanf("%d %d", &n, &m);
    for(i = 0; i < m; i++)
    {
        scanf("%s", s[i]);
        for(j = 0; j < n; j++)
        {
            x = 0;
            for(k = j + 1; k < n; k++)
            {
                if(s[i][j] > s[i][k])
                {
                    x++;
                }
            }
            a[i] += x;
        }
    }
    qs(0, m - 1);
    for(i = 0; i < m; i++)
    {
        printf("%s\n", s[i]);
    }
    return 0;
}
void qs(int l, int r)
{
    if(l >= r) return ;
    int i = l, j = r, x = a[l];
    char key[55];
    strcpy(key, s[l]); //快排，将大小快排的同时也将每一排换掉（没有用到结构体）
    while(i < j)
    {
        while(i < j && a[j] >= x)
        {
            j--;
        }
        a[i] = a[j];
        strcpy(s[i], s[j]);
        while(i < j && a[i] <= x)
        {
            i++;
        }
        a[j] = a[i];
        strcpy(s[j], s[i]);
    }
    a[i] = x;
    strcpy(s[i], key);
    qs(l, i - 1);
    qs(i + 1, r);
}