DNA Sorting
Time Limit: 2 Seconds Memory Limit: 65536 KB
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.
This problem contains multiple test cases!
The first line of a multiple input is an integer N, then a blank line followed by N input blocks. Each input block is in the format indicated in the problem description. There is a blank line between input blocks.
The output format consists of N output blocks. There is a blank line between output blocks.
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 (1 < 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''. If two or more strings are equally sorted, list them in the same order they are in the input file.
Sample Input
1
10 6
AACATGAAGG
TTTTGGCCAA
TTTGGCCAAA
GATCAGATTT
CCCGGGGGGA
ATCGATGCAT
Sample Output
1 #include <iostream>
2 #include <string>
3 #include <vector>
4 #include <algorithm>
5 using namespace std;
6 bool cmp(const string &s1, const string &s2){
7 int i, j;
8 int l1 = s1.length(), l2 = s2.length();
9 int c1 = 0, c2 = 0;
10 for(i = 0; i < l1 - 1; i++){
11 for(j = i + 1; j < l1; j++){
12 if(s1[i] > s1[j])
13 c1++;
14 }
15 }
16 for(i = 0; i < l2 - 1; i++){
17 for(j = i + 1; j < l2; j++){
18 if(s2[i] > s2[j])
19 c2++;
20 }
21 }
22 //如果逆序数相同，那么会按原先固有的位置排序
23 return c1 < c2;
24 }
25
26 int main(){
27 int t;
28 int n, m;
29 string str;
30 vector<string> v;
31 cin >> t;
32 while(t--){
33 cin >> n >> m;
34 v.clear();
35 for(int i = 0; i < m; i++){
36 cin >> str;
37 v.push_back(str);
38 }
39 sort(v.begin(), v.end(), cmp);
40 for(vector<string>::iterator it = v.begin(); it != v.end(); it++){
41 cout << *it << endl;
42 }
43 if(t != 0)
44 cout << endl;
45 }
46 //system("pause");
47 return 0;
48 } 越努力，越幸运