POJ 1007 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.
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
题目大致意思:很水的一道题,就是让你按照序列的大小从大到小重新排序一次就好,序列的大小就是第一个字母后面比它大的所有字母的个数+第二个字母后面比它大的所有字母的个数+......
代码如下:
#include<iostream> using namespace std; struct N { char a[50]; int num; }Line[100]; void quicksort(int left, int right) //快速排序 { int i, j; struct N temp; if (left < right) { i = left, j = right, temp = Line[i]; while (i < j) { while (Line[j].num <= temp.num&&i < j) j--; Line[i] = Line[j]; while (Line[i].num >= temp.num&&i < j) i++; Line[j] = Line[i]; } Line[i] = temp; quicksort(left, j - 1); quicksort(j + 1, right); } } int main() { int n, m; int count; cin >> m >> n; for (int i = 0; i < n; i++) for (int j = 0; j < m; j++) cin >> Line[i].a[j]; for (int i = 0; i < n; i++) { count = 0; for (int j = 0; j < m; j++) { char ch = Line[i].a[j]; for (int k = j; k < m; k++) if (Line[i].a[k] >= ch) count++; } Line[i].num = count; } quicksort(0, n-1); for (int i = 0; i < n; i++) cout << Line[i].a << endl; }