【POJ】1007 DNA Sorting
DNA Sorting
Time Limit: 1000MS
Memory Limit: 10000K
Total Submissions: 60701
Accepted: 23971
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.
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
Sample Output
Source
Time Limit: 1000MS
Memory Limit: 10000K
Total Submissions: 60701
Accepted: 23971
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.
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
#include
#include
#include
using namespace std;
typedef struct pair {
string s;
int inversion;
}Pair;
int calc_inversion(string s, int length)
{
int sum = 0;
int A = 0;
int C = 0;
int G = 0;
int T = 0;
// ,
for(int i = length - 1; i >= 0; --i) {
switch(s[i]) {
case 'A':
++A;
break;
case 'C':
sum += A;
++C;
break;
case 'G':
sum += A + C;
++G;
break;
case 'T':
sum += A + C + G;
++T;
break;
default:
break;
}
}
return sum;
}
bool comp(const Pair& p1, const Pair& p2) {
return p1.inversion <= p2.inversion;
}
int main()
{
int n, length;
cin >> length >> n;
vector vec;
for(int i = 0; i != n; ++i) {
Pair p;
cin >> p.s;
p.inversion = calc_inversion(p.s, length);
vec.push_back(p);
}
sort(vec.begin(), vec.end(), comp);
for(vector::iterator beg = vec.begin();
beg != vec.end();
++beg) {
cout << beg->s << endl;
}
}