UVA 10810 - Ultra-QuickSort（树状数组+离散化) - c++编程基础

Problem B: Ultra-QuickSort

In this problem, you have to analyze a particular sorting algorithm. The algorithm processes a sequence of ndistinct integers by swapping two adjacent sequence elements until the sequence is sorted in ascending order. For the input sequence

9 1 0 5 4 ,

Ultra-QuickSort produces the output

0 1 4 5 9 .

Your task is to determine how many swap operations Ultra-QuickSort needs to perform in order to sort a given input sequence.

The input contains several test cases. Every test case begins with a line that contains a single integer n < 500,000 -- the length of the input sequence. Each of the the following n lines contains a single integer 0 ≤ a[i] ≤ 999,999,999, the i-th input sequence element. Input is terminated by a sequence of length n = 0. This sequence must not be processed.

For every input sequence, your program prints a single line containing an integer number op, the minimum number of swap operations necessary to sort the given input sequence.

Sample Input

Output for Sample Input

6
0

题意：问排序好需要交换几次。

思路：利用树状数组求逆序对。。但是数字很大的时候就需要离散化。

代码：

#include 
  
   
#include 
   
     #include 
     #include 
     
       using namespace std; const int N = 500005; int n, num[N], save[N], bit[N]; map
      
        hash; void add(int x, int v) { while (x <= n) { bit[x] += v; x += (x&(-x)); } } int get(int x) { int ans = 0; while (x > 0) { ans += bit[x]; x -= (x&(-x)); } return ans; } void init() { hash.clear(); memset(bit, 0, sizeof(bit)); for (int i = 1; i <= n; i++) { scanf("%d", &num[i]); save[i] = num[i]; } sort(save + 1, save + n + 1); for (int j = 1; j <= n; j++) hash[save[j]] = j; } long long solve() { long long ans = 0; for (int i = 1; i <= n; i++) { ans += get(n) - get(hash[num[i]]); add(hash[num[i]], 1); } return ans; } int main() { while (~scanf("%d", &n) && n) { init(); printf("%lld\n", solve()); } return 0; }