poj1007——DNA Sorting
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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 6AACATGAAGGTTTTGGCCAATTTGGCCAAAGATCAGATTTCCCGGGGGGAATCGATGCAT
Sample Output
CCCGGGGGGAAACATGAAGGGATCAGATTTATCGATGCATTTTTGGCCAATTTGGCCAAA
题意大概是:设一个字母序列”DAABEC“的逆序数是5,因为D比它右边的4个字母大,而E比它右边的1个字母大。序列”AACEDGG“的逆序数是1,几乎已经排好序
现在对DNA字符串序列进行分类。然而,分类不是按字母顺序,而是按”排序“的次序,从最多已知排序到最少已知排序
#include<iostream>#include<cstdio>#include<algorithm>using namespace std;struct aa{ char s[55]; int x;};bool cmp(aa m,aa n){ return m.x<n.x;}int main(){ int m,n,i,j,k; char s[10]; cin>>n>>m; aa a[110]; gets(s); for(i=0; i<m; ++i) { gets(a[i].s); a[i].x=0; for(j=0; j<n; ++j) for(k=j+1; k<n; k++) { if(a[i].s[j]>a[i].s[k]) a[i].x++; } } sort(a,a+m,cmp); for(i=0;i<m;++i) cout<<a[i].s<<endl; return 0;}
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