DNA Sorting poj1007 解题报告

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

解题思路：这题主要要看懂题意就KO了，知道了就很简单了，刚开始一直不知道是 什么回事的，一直不知怎么出那个结果的，最后Google一下就完全明白了，只是暴力求 出逆序和，然后排序，没 什么可以说的，也没什么复杂的算法，只是简单的一些操作，就随便在网上copy了 一个代码解决问题了，但还是要学会 结构体排序等等。    代码如下： 

Code:

1. #include<iostream>
2. #include<algorithm>
3. #include<string>
5. usingnamespacestd;
7. structdna
8. {
9. intpos;
10. intkey;
11. stringstr;
12. };
14. /*sort比较函数*/
15. boolcmp(constdna&a,constdna&b)
16. {
17. if(a.key!=b.key)
18. {
19. returna.key<b.key;
20. }
21. else
22. {
23. returna.pos<b.pos;
24. }
25. }
27. intmain()
28. {
29. intn,m,count;
30. dnainv[110];
31. stringstr;
32. cin>>n>>m;
34. /*求逆序数对的个数*/
35. for(inti=0;i<m;i++)
36. {
37. cin>>str;
38. count=0;
39. for(intj=0;j<n-1;j++)
40. {
41. for(intk=j+1;k<n;k++)
42. {
43. if(str[j]>str[k])count++;
44. }
45. }
46. /*保存信息*/
47. inv[i].key=count;
48. inv[i].pos=i;
49. inv[i].str=str;
50. }
52. /*按逆序数对大小/序号排序*/
53. sort(inv,inv+m,cmp);
55. /*输出结果*/
56. for(inti=0;i<m;i++)
57. {
58. cout<<inv[i].str<<endl;
59. }
61. //system("pause");
62. return0;
63. }