POJ—DNA Sorting

最新推荐文章于 2019-01-22 10:18:06 发布

_鶸鸡

最新推荐文章于 2019-01-22 10:18:06 发布

阅读量212

点赞数

CC 4.0 BY-SA版权

分类专栏： 2017年10月

本文链接：https://blog.youkuaiyun.com/qq_35786972/article/details/78220610

2017年10月专栏收录该内容

6 篇文章

订阅专栏

本文详细介绍了DNASorting算法的基本原理及其实现过程。通过计算DNA序列的“未排序度”来衡量其排序程度，并以此为依据对DNA字符串进行排序。文章还分享了使用map和vector等数据结构解决该问题的经验。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

DNA Sorting

Time Limit: 1000MS Memory Limit: 10000K

Total Submissions: 103622 Accepted: 41507

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

终于做完的题目，长吁一口气。

1.通过这道题再学了map，懂了基本用法，终于完全懂了基本用法了（是学长教的，哈哈哈）

以前总觉得像隔着雾气氤氲的玻璃看，现在终倒是敲破了玻璃。

希望能以后能敲破越来越多的玻璃。

2.get到的新知识：vector<int>vec能保证没有相同的数据。

map<int,string> mp;WA//正因为可能出现相同的测试案例

map<vector<int>,string>//不会出现相同的测试案例，AC

附上代码：

#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<map>
#include<functional>
#include<vector>
using namespace std;

int main()
{
    int strn,strl;
    string str[110];
    int sortans[110];
    scanf("%d%d",&strl,&strn);
    map<vector<int>,string> mp;
    for(int i=0;i<strn;i++)
    {
        int cnt=0;
        cin>>str[i];
        vector<int>vec;
        for(int j=0; j<strl; j++)
        {
            for (int k = j; k<strl; k++)
            {
                if (str[i][j]>str[i][k])
                    cnt++;
            }
        }
        vec.push_back(cnt);
        vec.push_back(i);//判断是否有重复数据。
        mp[vec]=str[i];
    }
    for(map<vector<int>,string>::iterator it=mp.begin();it!=mp.end();it++)
        cout<<it->second<<endl;
    return 0;
}