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POJ 1007 DNA Sorting

By Sandra Sullivan,2014-12-01 11:51
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POJ 1007 DNA Sorting

POJ 1007 DNA Sorting

    简单题;暴力计算 32Ms

    DNA Sorting

    Time Limit: 1000MS Memory Limit: 10000K

    Total Submissions: 50312 Accepted: 19682

    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

下面贴代码;

    #include<iostream> #include<string>

    #include<algorithm> using namespace std; typedef struct

    {

     string str;

     int num; //逆序数

    }Node;

    int cmp(Node a, Node b) //sort 的比较函数 {

     return a.num < b.num; }

    int main()

    {

     int n, m, i, sum;

     Node node[100];

     while(scanf("%d", &n) != EOF) //控制输入

     {

     scanf("%d", &m);

     for(i = 0; i < m; i ++)

     cin >> node[i].str;

     for(i = 0; i < m; i ++)

     {

     sum = 0;

     for(int j = 0; j < n; j ++)

     {

     for(int k = j + 1; k < n; k ++) //计算逆序数

     {

     if(node[i].str[j] > node[i].str[k])

     sum ++;

     }

     }

     node[i].num = sum;

     }

     sort(node, node + m, cmp); //排序

     //输出 for(i = 0; i < m; i ++)

     cout << node[i].str << endl;

     }

     return 0;

    }

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