DNA Sorting
Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 102983 | Accepted: 41251 |
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
Source
用一个数组存原地址,一个数组存更改地址,遍历找到改掉后的地址,输出对应的字符串
#include<stdio.h> #include<string.h> #include<algorithm> using namespace std; int main() { char a[110][55]; int i, j, k; int dic[110], dic1[110]; int m, n; scanf("%d%d",&n,&m); memset(dic,0,sizeof(dic)); memset(dic1,0,sizeof(dic1)); for (i = 0; i < m; i++) scanf("%s",a[i]); for (i = 0; i < m; i++) for (j = 0; j < n - 1; j++) for (k = j + 1; k < n; k++) if (a[i][k] < a[i][j]) dic[i]++,dic1[i]++; sort(dic, dic + m); for (i = 0; i < m; i++) for (j = 0; j < m; j++) if (dic[i] == dic1[j]) { printf("%s\n", a[j]); dic1[j] = -1; break; } return 0; }