北大 ACM 1007 DNA Sorting

DNA Sorting

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 75079		Accepted: 30074

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

#include <iostream>

#include <map>

#include <string>

#include <vector>

#include <algorithm>

//#define MAXFORVALUE 1000 // 所能容纳的最大权值 

using namespace std;

typedef pair<string, int>  PAIR;

 

string *str=NULL;

map<string, int> result;

int count_num=0;

int size = 0;

string *repeat = new  string[50];

int repeat_num = 0;

 

void InputMessage();

void SortForValue();

void OutputMessage();

 

int main()

{

    InputMessage();

    SortForValue();

    OutputMessage();

 

    //int f;

    //cin >> f;

    return 1;

}

 

int cmp(const PAIR& x,  const PAIR& y)

{

    return x.second < y.second;

}

 

void OutputMessage()

{

    map<string, int>::const_iterator map_it = result.begin();

    vector<PAIR> vecpair;

 

    for(map<string, int>::iterator curr = result.begin(); curr != result.end(); ++curr)

    {

        vecpair.push_back(make_pair(curr->first, curr->second));

    }

    sort(vecpair.begin(), vecpair.end(), cmp);

    int count_repeat = 0;

    int j = 0;

    int flag = false;

    for(unsigned int i=0; i < vecpair.size(); i++)

    {

        for(int j = 0; j < repeat_num; j++)

        {

            if(vecpair[i].first == repeat[j])

            {

                cout << repeat[j] << endl;

            }

        }

        cout << vecpair[i].first << endl;

    }

}

 

void SortForValue()

{

    int value = 0; //得到字串的权值

     

    for(int i=0; i<count_num; i++)

    {

        for(int j=0; j < size; j++)

        {

            for(int k=j+1; k <size; k++)

            {

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

                {

                    value++;

                }

            }

        }

        pair<map<string, int>::iterator, bool> ret = 

            result.insert(make_pair(str[i], value));

        if(!ret.second) //如果出现重复的字串

        {

            repeat[repeat_num] = str[i];

            repeat_num++;

        }

        value = 0;

    }

}

 

void InputMessage()

{

    cin >> size >> count_num;

    str = new string[count_num];

    for(int i=0; i < count_num; i++)

        cin >> str[i];

}

其实说实话，这道题目挺简单的，我觉得关键在于细节的处理方面。可是我却调试了2天......经过这一件事情，我觉得不管是在编写代码还是调试代码时都要心静，要同盘考虑




很享受在通过ACM 时，出现的那个“Accepted”!