Description

In this problem, you have to analyze a particular sorting algorithm. The algorithm processes a sequence of n distinct 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.

5
9
1
0
5
4
3
1
2
3
0

6
0

題解：

樹狀數組求逆序數，先離散化，不然1e9開不下，然後每次對一個數id[i]，將樹狀數組中這個位置+1,統計這個位置的前綴和，然後用i-sum(id[i])(差實際上就是id[i]位置後綴和)就是這個數id[i]導致的逆序對數量。

代碼:

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<cstring>
 4 #include<algorithm>
 5 using namespace std;
 6 #define INF 0x3f3f3f3f
 7 #define M(a, b) memset(a, b, sizeof(a))
 8 const int N = 5e5 + 10;
 9 int c[N], id[N], n;
10 struct node {
11     int val, pos;
12     bool operator < (const node &rhs) const {
13         return val < rhs.val;
14     }
15 }a[N];
16 
17 int lowbit(int x) {return x & -x;}
18 
19 void add(int x, int d) {
20     while (x <= n) {
21         c[x] += d;
22         x += lowbit(x);
23     }
24 }
25 
26 int sum(int x) {
27     long long ret = 0;
28     while (x) {
29         ret += c[x];
30         x -= lowbit(x);
31     }
32     return ret;
33 }
34 
35 int main() {
36     while (scanf("%d", &n), n) {
37         M(c, 0);
38         for (int i = 1; i <= n; ++i)
39             scanf("%d", &a[i].val), a[i].pos = i;
40         sort(a+1, a+1+n);
41         for (int i = 1; i <= n; ++i) id[a[i].pos] = i;
42         long long ans = 0;
43         for (int i = 1; i <= n; ++i) {
44             add(id[i], 1);
45             ans += i-sum(id[i]);
46         }
47         printf("%lld\n", ans);
48     }
49 
50     return 0;
51 }

ZOJ-2386 Ultra-QuickSort 【樹狀數組求逆序數+離散化】