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  • POJ 1007 DNA Sorting

    链接:http://poj.org/problem?

    id=1007

    DNA Sorting

    Time Limit: 1000MS Memory Limit: 10000K
    Total Submissions: 89319 Accepted: 35892

    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


    Source

    East Central North America 1998

    大意——对于一个序列的无序性能够使用相互之间无序的元素组的个数表示。问:对于给定的多个DNA字符串(长度同样,由A, C, G和T组成),从最有序到最无序进行排列,而且输出。

    思路——定义一个结构体类型DNA。包括两个成员。一个str,表示DNA序列。一个count。表示逆序数。

    先计算出各DNA的逆序数,然后用sort函数对其按count进行从小到大排序,最后按排好序的数组进行输出就可以。

    复杂度分析——时间复杂度:O(m*n^2),空间复杂度:O(m*(m+n))

    附上AC代码:


    #include <iostream>
    #include <cstdio>
    #include <iomanip>
    #include <string>
    #include <cstring>
    #include <cmath>
    #include <algorithm>
    using namespace std;
    const double PI = acos(-1.0);
    const int MAX = 100;
    struct DNA
    {
        string str;
        int count;
    } dna[MAX];
    
    bool cmp(DNA a, DNA b);
    
    int main()
    {
        ios::sync_with_stdio(false);
        int n, m;
        while (cin >> n >> m)
        {
            for (int i=0; i<m; i++)
            {
                cin >> dna[i].str;
                dna[i].count = 0;
            }
            for (int i=0; i<m; i++)
            {
                for (int j=0; j<n; j++)
                    for (int k=j+1; k<n; k++)
                        if (dna[i].str[j] > dna[i].str[k])
                            dna[i].count++;
            }
            sort(dna, dna+m, cmp);
            for (int i=0; i<m; i++)
                cout << dna[i].str << endl;
        }
        return 0;
    }
    
    bool cmp(DNA a, DNA b)
    {
        return a.count < b.count;
    }
    


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  • 原文地址:https://www.cnblogs.com/yfceshi/p/6881871.html
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