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

这道题只要读懂题意其实还是很简单的。。大概意思就是说，题目中定义了一种排序方法，就是每个字符串中，按字典序，右边的字母应该比左边的字母大，那么我们就可以从左到右依次扫描每个字母，然后再从这个字母开始向右扫描有没有比它小的字母，如果有给当前字符串的“未排序程度”（暂时这么叫）加上1。
最后，按照这个未排序程度从小到大输出原字符串即可。

程序：

#include <stdio.h>/*poj 1007 DNA Sorting 2011-3-6*/#define MAXNUM 10000000char data[110][60];int sort[110];int n,m;int main(){int i,j,k,min,print;scanf("%d%d",&n,&m);for(i=1;i<=m;i++){scanf("%s",data[i]);for(j=0;j<n;j++)for(k=j+1;k<n;k++)if(data[i][j]>data[i][k])sort[i]++;}for(i=1;i<=m;i++){min=MAXNUM;for(j=1;j<=m;j++)if(sort[j]<min){print=j;min=sort[j];}sort[print]=MAXNUM;printf("%s/n",data[print]);}return 0;}