DNA Sorting
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 83442 | Accepted: 33584 |
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.
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 6 AACATGAAGG TTTTGGCCAA TTTGGCCAAA GATCAGATTT CCCGGGGGGA ATCGATGCAT
Sample Output
CCCGGGGGGA AACATGAAGG GATCAGATTT ATCGATGCAT TTTTGGCCAA TTTGGCCAAA
重点，归并排序求逆序数
1 #include <stdio.h>
2 #include <malloc.h>
3 #include <string.h>
4 #define MAXN 256
5 char a[MAXN];
6 char c[MAXN];
7 int cnt=0;
8 void MergeSort(int l, int r){
9 int mid, i, j, tmp;
10 if( r > l+1 ){
11 mid = (l+r)/2;
12 MergeSort(l, mid);
13 MergeSort(mid, r);
14 tmp = l;
15 for( i=l, j=mid; i < mid && j < r; ){
16 if( a[i] > a[j] ){
17 c[tmp++] = a[j++];
18 cnt += mid-i; //
19 }
20 else c[tmp++] = a[i++];
21 }
22 if( j < r ) for( ; j < r; ++j ) c[tmp++] = a[j];
23 else for( ; i < mid; ++i ) c[tmp++] = a[i];
24 for ( i=l; i < r; ++i ) a[i] = c[i];
25 }
26 }
27 int main(void){
28 int n,m;
29 scanf("%d%d",&n,&m);
30 char ** strs = (char **)malloc(m*sizeof(char*));
31 int * cnts = (int *)malloc(m*sizeof(int));
32 int i;
33 for(i = 0;i<m;i++){
34 strs[i] = (char *)malloc((n+1)*sizeof(char));
35 scanf("%s",strs[i]);
36 //printf("%s
",strs[i]);
37 cnt = 0;
38 strcpy(a,strs[i]);
39 //printf("%s
",a);
40 MergeSort(0,n);
41 //printf("%d
",cnt);
42 cnts[i] = cnt;
43 }
44
45 for(i = 0;i<m-1;i++){
46 int j,p=i;
47 for(j = i+1;j<m;j++){
48 if(cnts[p]>cnts[j]){
49 p = j;
50 }
51 }
52 if(p!=i){
53 int tmp = cnts[p];
54 cnts[p] = cnts[i];
55 cnts[i] = tmp;
56 char * str = strs[p];
57 strs[p] = strs[i];
58 strs[i] = str;
59 }
60 }
61 for(i = 0;i<m;i++){
62 printf("%s
",strs[i]);
63 free(strs[i]);
64 }
65 free(strs);
66 free(cnts);
67 return 0;
68 }