【POJ】1007 DNA Sorting

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 6AACATGAAGGTTTTGGCCAATTTGGCCAAAGATCAGATTTCCCGGGGGGAATCGATGCAT

Sample Output

CCCGGGGGGAAACATGAAGGGATCAGATTTATCGATGCATTTTTGGCCAATTTGGCCAAA

这个题目的意思并不是求DNA序列的ATCG配对，

而是求这些序列的稳定程度由低到高排序。

稳定程度指的是这个字符数组中的逆序数之和。

知道这个之后就很简单了。

要注意的是，排序的时候调用qsort或者sort函数都可以，

冒泡排序的话可能会TLE。

怒贴代码：

#include<stdio.h>#include<iostream>#include<stdlib.h>#include<algorithm>using namespace std;struct DNA{    char atcg[55];    int cas;};DNA dna[110];bool cmp(DNA a,DNA b){    return a.cas<b.cas;}int main(){    int n,m;    while(scanf("%d%d",&n,&m)!=EOF)    {        for(int i=0;i<m;i++)        {            scanf("%s",dna[i].atcg);            dna[i].cas=0;            for(int j=0;j<n-1;j++)                for(int k=j+1;k<n;k++)                {                    if(dna[i].atcg[j]>dna[i].atcg[k])                        dna[i].cas++;                }        }        sort(dna,dna+m,cmp);        for(int i=0;i<m;i++)        {            printf("%s\n",dna[i].atcg);        }    }    return 0;}