POJ

DNA Sorting

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 101498 Accepted: 40675

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排序
#include<iostream>#include<algorithm>using namespace std;struct AP{    int x;    char c[108];}a[108];bool compare(AP a,AP b);int main(){    int n,m;    while(cin>>n>>m!=NULL)    {        for(int i=0;i<m;i++)        {            cin>>a[i].c;            a[i].x=0;            for(int j=0;j<(n-1);j++)                for(int k=j+1;k<n;k++)                    if(a[i].c[j]>a[i].c[k])                        a[i].x++;        }        sort(a,a+m,compare);        for(int i=0;i<m;i++)            cout<<a[i].c<<endl;    }    return 0;}bool compare(AP a,AP b){    return a.x<b.x;}