poj1007 - DNA Sorting
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2022-03-13 22:47:02
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DNA Sorting
Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 76944 | Accepted: 30826 |
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
利用string来存储DNA序列然后利用其逆序数从小到大进行排序,输出就行了
下面是我的源代码:
#include<iostream> #include<cstring> #include<algorithm> using namespace std; int cal(string s) { int sum=0,i,j,len; len=s.size(); for(i=0;i<len-1;i++) for(j=i+1;j<len;j++) if(s[i]>s[j]) sum+=1; return sum; } bool cmp(string const &s1,string const &s2) { return cal(s1)<cal(s2); } int main() { string str[105]; int n,m,i; cin>>n>>m; for(i=0;i<m;i++) cin>>str[i]; sort(str,str+m,cmp); for(i=0;i<m;i++) cout<<str[i]<<endl; return 0; }
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