【POJ】1007 DNA Sorting

DNA Sorting

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 60701		Accepted: 23971

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

Source

#include
#include
#include
#include

using namespace std;

typedef struct pair {
	string s;
	int inversion;
}Pair;

int calc_inversion(string s, int length) 
{
	int sum = 0;
	int A = 0;
	int C = 0;
	int G = 0;
	int T = 0;
        // 从后向前，一次遍历
	for(int i = length - 1; i >= 0; --i) {
		switch(s[i]) {
		case 'A': 
			++A;
			break;
		case 'C':
			sum += A;
			++C;
			break;
		case 'G':
			sum += A + C;
			++G;
			break;
		case 'T':
			sum += A + C + G;
			++T;
			break;
		default:
			break;
		}
	}
	return sum;
}

bool comp(const Pair& p1, const Pair& p2) {
	return p1.inversion <= p2.inversion; 
}

int main() 
{
	int n, length;

	cin >> length >> n;
	vector vec;
	for(int i = 0; i != n; ++i) {
		Pair p;
		cin >> p.s;
		p.inversion = calc_inversion(p.s, length);
		vec.push_back(p);
	}
	sort(vec.begin(), vec.end(), comp);
	for(vector::iterator beg = vec.begin();
		beg != vec.end();
		++beg) {
			cout << beg->s << endl;
	}

}