poj 1007 DNA Sorting(排序--快排)
DNA Sorting
Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 67603 | Accepted: 26858 |
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
#include <iostream> #include <algorithm> using namespace std; typedef struct { string dna; int count; }DNA; DNA dna[101]; int cmp(const void *a,const void *b) { DNA *aa = (DNA *)a; DNA *bb = (DNA *)b; return aa->count-bb->count; } int main() { int n,m; char c; cin>>n>>m; for(int i = 0; i < m; i++) { cin>>dna[i].dna; dna[i].count = 0; for(int j = 0; j < n; j++) for(int k = j+1; k < n; k++) { if(dna[i].dna[j]>dna[i].dna[k]) dna[i].count++; } } qsort(dna,m,sizeof(dna[0]),cmp); for(int i = 0; i < m; i++) cout<<dna[i].dna<<endl; return 0; }
posted on 2012-09-26 19:00 NewPanderKing 阅读(2820) 评论(0) 编辑 收藏 举报