poj1007

DNA Sorting

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 89161 Accepted: 35831

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

恩，题目大意就是给两个数n和m，n代表字符串长度，m代表字符串个数。恩，对于每个字符串来说，当前字符之后有多少个字符比它小按字典序排列，然后所有字符的都加在一起表示为该字符串的等级，然后按照等级从小到大排列输出即可。

啊，水题，结构体sort排序就ok了，当时没做竟然是因为没看懂题，英语不好是硬伤啊。。。。

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;struct node{int c;char str[101];}ss[500];int cmp(node a,node b){return a.c<b.c;}int main(){int n,m,i,j,len;while(~scanf("%d %d",&n,&m)){for(i=0;i<m;++i){scanf("%s",ss[i].str);len=strlen(ss[i].str);ss[i].c=0;for(j=0;j<len;++j){for(int k=j+1;k<len;++k){if(ss[i].str[k]<ss[i].str[j])ss[i].c++;}}}sort(ss,ss+m,cmp);for(i=0;i<m;++i)    printf("%s\n",ss[i].str);}return 0;}