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

10 6
AACATGAAGG
TTTTGGCCAA
TTTGGCCAAA
GATCAGATTT
CCCGGGGGGA
ATCGATGCAT
Sample Output

CCCGGGGGGA
AACATGAAGG
GATCAGATTT
ATCGATGCAT
TTTTGGCCAA
TTTGGCCAAA

原题连接：点击打开链接

题意：给出DNA序列，求解每条序列的逆序对数，并按逆序对数从小到大排列。关键是如何求解逆序对数。

这里用的是二叉树。二叉树节点见代码的TreeNode，same表示跟该节点值相等的个数，初始值为1，small表示比该节点值小的值的个数。从每条DNA的最后一个节点开始，从后向前插入到二叉树中。然后在排序。

代码如下：

#include<iostream>#include<string>#include<algorithm>#include<memory.h>#include<limits.h>using namespace std;int res = 0;struct TreeNode {    char val;    int small;    int same;    TreeNode *left,*right;    TreeNode(char v) {        val = v;        same = 1;        small = 0;        left = right = NULL;    }};struct Node {    string str;    int cnt;};TreeNode* insertTree(TreeNode *root, char key) {    if(root == NULL) {        root = new TreeNode(key);    }    else if(root->val == key) {        res += root->small;        root->same += 1;    }    else if(root->val < key) {        res += root->small + root->same;        root->right = insertTree(root->right,key);    }    else if(root->val > key) {        root->small += 1;        root->left = insertTree(root->left, key);    }    return root;}bool cmp(const Node &a,const Node &b) {    return a.cnt < b.cnt;}int main() {    Node arr[105];    int m,n;    cin>>m>>n;    for(int i = 0; i < n; i++) {        res = 0;        string s;        cin>>s;        arr[i].str = s;        TreeNode *root = NULL;        for(int j = m-1; j >= 0; j--) {            root = insertTree(root,s[j]);        }        arr[i].cnt = res;    }    sort(arr,arr+n,cmp);    for(int i = 0; i < n; i++) {        cout<<arr[i].str<<endl;    }    return 0;}