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”!