浙大在线评测系统 1188 DNA Sorting

Problem:

    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.

    This problem contains multiple test cases!

    The first line of a multiple input is an integer N, then a blank line followed by N input blocks. Each input block is in the format indicated in the problem description. There is a blank line between input blocks.

    The output format consists of N output blocks. There is a blank line between output blocks.


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 (1 < m <= 100) giving the number of strings. These are followed by m lines, each containing a string of length


Output:

    Output the list of input strings, arranged from ``most sorted'' to ``least sorted''. If two or more strings are equally sorted, list them in the same order they are in the input file.


Sample Input:

1

10 6
AACATGAAGG
TTTTGGCCAA
TTTGGCCAAA
GATCAGATTT
CCCGGGGGGA
ATCGATGCAT


Sample Output:

CCCGGGGGGA
AACATGAAGG
GATCAGATTT
ATCGATGCAT
TTTTGGCCAA
TTTGGCCAAA


Solution:

#include <iostream>
#include <string>
#include <map>
using namespace std;

// 声明：本代码仅供学习之用，请不要作为个人的成绩提交。
// http://blog.csdn.net/mskia
// email: [email protected]

int main( void ) {
    int n;
    cin >> n;
   
    for ( int i = 0; i < n; ++i ) {
        int len , row;
        cin >> len >> row;

        multimap< int , string > dnas;
        for ( int j = 0; j < row; ++j ) {
            string seq;
            cin >> seq;
           
            int ca = 0 , cc  = 0 , cg = 0 , total = 0;
            for ( int k = len - 1; k >= 0; --k ) {
                switch ( seq[k] ) {
                    case 'T':
                        total += cg + cc + ca;
                        break;
                       
                    case 'G':
                        total += ca + cc;
                        ++cg;
                        break;
                       
                    case 'C':
                        total += ca;
                        ++cc;
                        break;
                       
                    case 'A':
                        ++ca;
                        break;
                }
            }
            dnas.insert( pair< int , string >( total , seq ) );
        }
       
        if ( i != 0 ) cout << endl;

        for ( map< int , string >::iterator p = dnas.begin( ); p != dnas.end(); ++p ) {
            cout << p->second << endl;
        }

    }

    return 0;
}

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