POJ1007,DNA Sorting，排序水题

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 6

AACATGAAGG

TTTTGGCCAA

TTTGGCCAAA

GATCAGATTT

CCCGGGGGGA

ATCGATGCAT

Sample Output

CCCGGGGGGA

AACATGAAGG

GATCAGATTT

ATCGATGCAT

TTTTGGCCAA

TTTGGCCAAA

分析：

一道很水的排序题，主要是按题目给出的乱序规则给每个串增加一个乱序属性值。

用Inv函数算出，之后排序就可以了。

code:

#include<iostream>
#include<cstdio>
#define MAX_LEN 50
#define MAX_NUM 100
using namespace std;
char DNA[MAX_NUM+5][MAX_LEN+5];
int inversion[MAX_NUM+5];
int order[MAX_NUM+5];
int Inv(char *s,int len)
{
int count=0,i,j;
for(i=0;i<len;i++)
for(j=i+1;j<len;j++)
{
if(s[i]>s[j]) count++;
}
return count;
}
int main()
{
int len,n,i,j;
cin>>len>>n;
getchar();
for(i=0;i<n;i++)
{
gets(DNA[i]);
inversion[i+1]=Inv(DNA[i],len);
}
inversion[0]=10000;
order[0]=0;
for(i=1;i<=n;i++)
{
order[i]=i;
for(j=1;j<=n;j++)
if(inversion[j]>inversion[order[i]]&&inversion[order[i]]<inversion[order[i-1]])
order[i]=j;
}
for(i=0;i<n;i++)
puts(DNA[order[i+1]]);
return 0;
}