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 6AACATGAAGGTTTTGGCCAATTTGGCCAAAGATCAGATTTCCCGGGGGGAATCGATGCAT
Sample Output
CCCGGGGGGAAACATGAAGGGATCAGATTTATCGATGCATTTTTGGCCAATTTGGCCAAA 求一串字符的逆序，并按逆序数从小到大的顺序排列
采用结构体 #include #include #include using namespace std;struct node{ int xh;//分别求逆序数，最后排序 string str;};struct node ans[110];int cmp(struct node n1,struct node n2){ return n1.xh < n2.xh;}int main(){ int n,m,t; cin >> n >> m; for(int i = 0; i < m; i++) { t = 0; cin >> ans[i].str; for(int j = 0; j < n-1; j ++) { for(int k = j; k < n;k++) { if(ans[i].str[j] > ans[i].str[k]) t++; } } ans[i].xh = t; } sort(ans,ans+m,cmp); for(int i = 0;i < m;i++) { cout << ans[i].str << endl; }}