POJ 1007 DNA Sorting

DNA Sorting

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 85118		Accepted: 34274

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

解题思路：
这道题也是属于简单题，基本思路就是先算出每个DNA序列的逆序数，然后排个序输出（注意这里需要用稳定排序，我用了C++库的qsort，简单定义一下不处理‘==’的情况就可以了）。
求逆序数稍微有点技巧

代码：

#include <iostream>
#include <algorithm>

/**************************************************
 * POJ 1007 DNA Sorting
 * author:樊列龙
 * 2014-12-16
 *************************************************/
typedef struct DNA
{
    int count;// 逆序数的个数
    char DNAStr[110];
} DNA;

int N;
int SIZE;
int countA,countC,countG;


/** 计算DNA字符串的逆序数  */
int getDNAInversionNumber(char* dnaStr)
{
    int count = 0;
    countA = countC = countG = 0;
    for(int j = SIZE-1; j >= 0; j--)
    {
        switch(dnaStr[j])
        {
        case 'A':
            countA++;
            break;
        case 'C':
            countC++;
            count += countA;
            break;
        case 'G':
            countG++;
            count += countA;
            count += countC;
            break;
        case 'T':
            count += countA;
            count += countC;
            count += countG;
            break;
        }
    }
    return count;
}

int cmp(const void* a, const void* b)
{
    DNA* DNA_A = (DNA*)a;
    DNA* DNA_B = (DNA*)b;
    return (DNA_A->count) - (DNA_B->count);
}

int main()
{
    using namespace std;
    while(cin >> SIZE >> N)
    {
        DNA* dnas = new DNA[N];
        for(int i = 0; i < N; i++)
        {
            cin >> dnas[i].DNAStr;
            dnas[i].count = getDNAInversionNumber(dnas[i].DNAStr);
        }

        qsort(dnas, N, sizeof(DNA), cmp);

        for(int i = 0; i < N; i++)
        {
            cout << dnas[i].DNAStr << endl;
        }
    }
    return 0;
}