poj-1007 简单排序

DNA Sorting

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 71751		Accepted: 28638

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

 

翻译：

序列“未排序程度”的一个计算方式是元素乱序的元素对个数。例如：在单词序列“DAABEC'”中，因为D大于右边四个单词，E大于C，所以计算结果为5。这种计算方法称为序列的逆序数。序列“AACEDGG”逆序数为1（E与D）——近似排序，而序列``ZWQM'' 逆序数为6（它是已排序序列的反序）。

你的任务是分类DNA字符串（只有ACGT四个字符）。但是你分类它们的方法不是字典序，而是逆序数，排序程度从好到差。所有字符串长度相同。

输入：

第一行包含两个数：一个正整数n（0<n<=50）表示字符串长度，一个正整数m（0<m<=100）表示字符串个数。接下来m行，每行一个长度为n的字符串。

输出：

输出输入字符串列表，按排序程度从好到差。如果逆序数相同，就原来顺序输出。

样例输入：

10 6
AACATGAAGG
TTTTGGCCAA
TTTGGCCAAA
GATCAGATTT
CCCGGGGGGA
ATCGATGCAT

样例输出：

CCCGGGGGGA
AACATGAAGG
GATCAGATTT
ATCGATGCAT
TTTTGGCCAA
TTTGGCCAAA

解题时间：2013/3/31

 

源代码：

 

#include <stdio.h>
#include <stdlib.h>

int main()
{
int n,length;
char str[100][51];
int L[100];//str
反序列个数
int pt[100];//下标
int i,j;
scanf("%d %d",&length,&n);
for(i=0;i<n;i++)
{
L[i]=0;
scanf("%s",str[i]);
int k;
for(j=0;j<length-1;j++)
{
for(k=j+1;k<length;k++)
{
if(str[i][k]<str[i][j])
L[i]++;
}
}
pt[i]=i;
}

//冒泡排序(稳定)
int temp;
for(i=1;i<n;i++)
{
for(j=0;j<n-i;j++)
{
if(L[pt[j]]>L[pt[j+1]])
{
temp=pt[j];
pt[j]=pt[j+1];
pt[j+1]=temp;
}
}
}
//输出
for(i=0;i<n;i++)
{
printf("%s\n",str[pt[i]]);
}
return 0;
}