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浙大在线评测系统 1188 DNA Sorting

王朝other·作者佚名  2006-01-09
窄屏简体版  字體: |||超大  

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: bitrain@hotmail.com

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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