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.
InputThe 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.
Sample Input
10 6
AACATGAAGG
TTTTGGCCAA
TTTGGCCAAA
GATCAGATTT
CCCGGGGGGA
ATCGATGCAT
Sample Output
CCCGGGGGGA
AACATGAAGG
GATCAGATTT
ATCGATGCAT
TTTTGGCCAA
TTTGGCCAAA
分析：
本题难点在于逆序数计算。惯性思维容易使人对字符串排序后进行比较，此方法复杂易错。
简单的方法是数一数每个字符右边有几个比它大的字符。
code：
1 #include <iostream>
2 #include <
string>
3 using namespace std;
4 int len;
5 int measureFunc(
string s);
6 int main()
7 {
8 int N;
9 int measure[
100];
10 string ss[
100];
11 cin>>len>>
N;
12 cin.ignore();
13 for(
int i=
0;i<N;i++
)
14 {
15 cin>>
ss[i];
16 }
17 for(
int i=
0;i<N;i++
)
18 {
19 measure[i]=
measureFunc(ss[i]);
20 }
21 int index=
0;
22 for(
int i=
0;i<N;i++
)
23 {
24 index=
0;
25 for(
int j=
0;j<N;j++
)
26 {
27 if(measure[index]>
measure[j])
28 index=
j;
29 }
30 cout<<ss[index]<<
endl;
31 measure[index]=
65536;//置一个大数
32 }
33
34 return 0;
35 }
36 int measureFunc(
string s)
37 {
38 int mea=
0;
39 for(
int i=
0;i<len;i++)
//计算字符串的逆序数
40 for(
int m=i+
1;m<len;m++
)
41 if(s[i]>
s[m])
42 mea++
;
43 return mea;
44 }
转载于:https://www.cnblogs.com/QuentinYo/archive/2013/04/01/2994329.html
