poj1007

DNA Sorting

Time Limit: 1000MS		Memory Limit: 10000K
Total 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 6
AACATGAAGG
TTTTGGCCAA
TTTGGCCAAA
GATCAGATTT
CCCGGGGGGA
ATCGATGCAT

Sample Output

CCCGGGGGGA
AACATGAAGG
GATCAGATTT
ATCGATGCAT
TTTTGGCCAA
TTTGGCCAAA

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;
}

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