POJ1007 DNA Sorting
2015-01-05 13:00
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DNA Sorting
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
East Central North America 1998
Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 85026 | Accepted: 34238 |
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
East Central North America 1998
#include <stdio.h> #include <string.h> #include <algorithm> using std::sort; #define maxn 102 struct Node { int inverNum; char str[52]; } arr[maxn]; int N, M; bool cmp(Node a, Node b) { if(a.inverNum == b.inverNum) return false; return a.inverNum < b.inverNum; } int getInver(char *str) { int sum = 0, i, j; for(i = 0; i < N - 1; ++i) for(j = i + 1; j < N; ++j) if(str[i] > str[j]) ++sum; return sum; } int main() { // freopen("stdin.txt", "r", stdin); int i, j; while(scanf("%d%d", &N, &M) == 2) { for (i = 0; i < M; ++i) { scanf("%s", arr[i].str); arr[i].inverNum = getInver(arr[i].str); } sort(arr, arr + M, cmp); for (i = 0; i < M; ++i) puts(arr[i].str); } return 0; }
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