DNA Sorting
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2185 Accepted Submission(s): 1064
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
Source
East Central North America 1998
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//输入块之间有空行， 输出块之间也有空行；
1 #include <stdio.h>
2 #include <string.h>
3 #include <algorithm>
4 using namespace std;
5
6 struct ac
7 {
8 int s;
9 char str[55];
10 };
11 ac num[110];
12
13 bool cmp(ac a, ac str)
14 {
15 return a.s < str.s;
16 }
17
18 int main()
19 {
20 char ch[55];
21 int sum, total = 0;
22 int t, i, j, k, m, n;
23 scanf("%d", &t);
24 while(t--)
25 {
26 if(total != 0)
27 printf("\n");
28 scanf("%d %d", &n, &m);
29 total++ ;
30 for(i=0; i<m; i++)
31 {
32 sum = 0;
33 scanf("%s", ch);
34 //puts(ch);
35 strcpy(num[i].str, ch);
36 //puts(num[i].str);
37 for(k=0; k<n-1; k++)
38 {
39 for(j=k+1; j<n; j++)
40 {
41 if(ch[k] > ch[j])
42 sum++;
43 //printf("%d %d\n", k, sum);
44 }
45 }
46 num[i].s = sum;
47 }
48 sort(num, num+m, cmp);
49 for(i=0; i<m; i++)
50 printf("%s\n", num[i].str);
51 if(t!=0)
52 printf("\n");
53 }
54 return 0;
55 }