POJ1007浅析------DNA Sorting（排列DNA）

POJ1007

 

DNA Sorting

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 64552		Accepted: 25498

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

Source

East Central North America 1998

 

 

题目类型：排序

 

 

题目大意：

        一个序列的“反序数”是指这个序列中未规则地排序（从小大大排序）的项的数量和。例如：“DAABEC”，它的度便是5，因为D比排在它后面字母中的4个大，E比排在它后面的1个字母大，所以它的反序数为4+1=5.

       现在给你 m 个长度为 n 的序列，按照各序列反序数从小到大顺序输出这些已排序数列。

       解题思路：

       一：先把这些序列输入，存放到二维字符型数组里；

       二：计算各序列反序数，存放到一个整型数组中；（前两步可以合在一起，输入的同时进行处理）

       三：按照反序数数组的大小排序；

       四：按照已有顺序输出各序列。

       由于对序列各个体都排序会比较难入手，比较容易想到的是把它的地址（对应的 i 值）存放到另一整型数组中，在反序数数组排序的同时进行排序。这样，按照反序数从小到大排列好的同时，与其所关联的序列一一对应。按照这个顺序，依次输出即可。

 

 

我的代码如下：

#include<stdio.h>
#include<string.h>
int main()
{
    int n,m,i,j,k,sum,temp;
    char str[105][55];
    int a[100],b[100];
    scanf("%d%d",&n,&m);
    for(i=0;i<m;i++)
    {
        scanf("%s",str[i]);
        sum=0;
        for(j=0;j<strlen(str[i])-1;j++)
        {
            for(k=j+1;k<strlen(str[i]);k++)
            {
                if(str[i][k]<str[i][j])
                    sum++;                           
            }                         
        }               
        a[i]=sum;
        b[i]=i;
    }
    for(i=0;i<m-1;i++)
    {
        for(j=0;j<m-1-i;j++)
        {
            if(a[j]>a[j+1])
            {
                temp=a[j];a[j]=a[j+1];a[j+1]=temp;
                temp=b[j];b[j]=b[j+1];b[j+1]=temp;
            }                    
        }                  
    }
    for(i=0;i<m;i++)
        printf("%s\n",str[b[i]]);   
    return 0;    
}

这题确实还是很水。。。排序的过程中采用了模拟的方法（知识个人见解）。。。