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 6AACATGAAGGTTTTGGCCAATTTGGCCAAAGATCAGATTTCCCGGGGGGAATCGATGCAT

Sample Output

CCCGGGGGGAAACATGAAGGGATCAGATTTATCGATGCATTTTTGGCCAATTTGGCCAAA

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>  
4.    
5. using namespace std;  
6.    
7. struct dna  
8. {  
9.     int pos;  
10.     int key;  
11.     string str;  
12. };  
13.    
14. /* sort比较函数 */  
15. bool cmp(const dna &a, const dna &b)  
16. {  
17.     if (a.key != b.key)  
18.     {  
19.         return a.key < b.key;  
20.     }  
21.     else  
22.     {  
23.         return a.pos < b.pos;  
24.     }  
25. }  
26.    
27. int main()  
28. {  
29.     int n, m, count;  
30.     dna inv[110];  
31.     string str;  
32.     cin >> n >> m;  
33.    
34.     /* 求逆序数对的个数 */  
35.     for (int i = 0; i < m; i++)  
36.     {  
37.         cin >> str;  
38.         count = 0;  
39.         for (int j = 0; j < n - 1; j++)  
40.         {  
41.             for (int k = 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.     }  
51.    
52.     /* 按逆序数对大小/序号排序 */  
53.     sort(inv, inv + m, cmp);  
54.    
55.     /* 输出结果 */  
56.     for (int i = 0; i < m; i++)  
57.     {  
58.         cout << inv[i].str << endl;  
59.     }  
60.    
61.     //system("pause");  
62.     return 0;  
63. }