DNA Sorting
Problem 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.
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
CCCGGGGGGA
AACATGAAGG
GATCAGATTT
ATCGATGCAT
TTTTGGCCAA
TTTGGCCAAA
此题傻暴力求逆序数即可，我用的归并排序求。
代码：
1 #include<cstdio>
2 #include<cstring>
3 #include<malloc.h>
4 #include<algorithm>
5 using namespace std;
6
7 int ans;
8
9 struct str
10 {
11 char s[105];
12 char sort_s[105];
13 int Num;
14 };
15
16 bool cmp(struct str a,struct str b)
17 {
18 return a.Num<b.Num;
19 }
20
21 void MergeArray(char a[],int L,int Mid,int R,char temp[])
22 {
23 int k=0;
24 int i=L,j=Mid+1;
25 int m=Mid,n=R;
26 while(i<=m&&j<=n)
27 {
28 if(a[i]<=a[j])
29 temp[k++]=a[i++];
30 else
31 {
32 temp[k++]=a[j++];
33 ans+=m-i+1;
34 }
35 }
36 while(i<=m)
37 temp[k++]=a[i++];
38 while(j<=n)
39 temp[k++]=a[j++];
40 for(i=0;i<k;i++)
41 a[L+i]=temp[i];
42 }
43
44 void MergeSort(char a[],int L,int R,char temp[])
45 {
46 if(L<R)
47 {
48 int Mid=(L+R)/2;
49 MergeSort(a,L,Mid,temp);
50 MergeSort(a,Mid+1,R,temp);
51 MergeArray(a,L,Mid,R,temp);
52 }
53 }
54
55 int main()
56 {
57 struct str S[55];
58 int i,m,n,t;
59 scanf("%d",&t);
60 while(t--)
61 {
62 scanf("%d%d",&m,&n);
63 for(i=0;i<n;i++)
64 scanf("%s",S[i].s);
65 for(i=0;i<n;i++)
66 {
67 ans=0;
68 int len=strlen(S[i].s);
69 strcpy(S[i].sort_s,S[i].s);
70 char *p=(char*)malloc(sizeof(char)*(len+1));
71 MergeSort(S[i].sort_s,0,len-1,p);
72 S[i].Num=ans;
73 free(p);
74 }
75 sort(S,S+n,cmp);
76 for(i=0;i<n;i++)
77 printf("%s\n",S[i].s);
78 }
79 return 0;
80 }