HDU 2689 sort it-from lanshui_Yang

Problem Description
You want to processe a sequence of n distinct integers by swapping two adjacent sequence elements until the sequence is sorted in ascending order. Then how many times it need.
For example, 1 2 3 5 4, we only need one operation: swap 5 and 4.

 

Input
The input consists of a number of test cases. each case consists of two lines: the first line contains a positive integer n (n <= 1000); the next line contains a permutation of the n integers from 1 to n.

Output
For each case, output the minimum times need to sort it in ascending order on a single line.

Sample Input
3
1 2 3
4
4 3 2 1

Sample Output
0
6. Give you a number of n, and then 1 ~ A permutation of n, allowing you to find the number of reverse orders in this arrangement. Solution: You can use a tree array to solve this problem. The tree array has three purposes: 1. Single Point update, interval SUM 2, Interval Update, single point sum 3, and reverse order number. The idea of Reverse Order is relatively simple. Please refer to the code

# Include <iostream> # include <cstring> # include <string> # include <cstdio> # include <cmath> # include <algorithm> # include <queue> using namespace std; const int MAXN = 1e5 + 7; int C [MAXN]; int n; int lowbit (int x) {return x &-x;} int sum (int x) {int sumt = 0; while (x> 0) {sumt + = C [x]; x-= lowbit (x);} return sumt;} void add (int x, int d) {while (x <= n) {C [x] + = d; x + = lowbit (x) ;}} int main () {whi Le (scanf ("% d", & n )! = EOF) {int I; int ans = 0; memset (C, 0, sizeof (C); for (I = 1; I <= n; I ++) {int a; scanf ("% d", & a); add (a, 1); // This is the essence of the entire program, please understand ans + = I-sum (a); // count the number of reverse orders} printf ("% d \ n", ans);} return 0 ;}