POJ——T 2299 Ultra-QuickSort

http://poj.org/problem?id=2299

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.

Input

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.

Output

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

5
9
1
0
5
4
3
1
2
3
0

Sample Output

6
0

Source

Waterloo local 2005.02.05

树状数组+离散化求逆序对

 1 #include <algorithm>
 2 #include <cstring>
 3 #include <cstdio>
 4
 5 using namespace std;
 6
 7 const int N(500000);
 8 long long ans;
 9 int n,tr[N];
10 struct Node
11 {
12     int num,mark;
13 }a[N];
14 bool cmp(Node a,Node b)
15 {
16     if(a.num==b.num)
17         return a.mark>b.mark;
18     return a.num>b.num;
19 }
20
21 #define lowbit(x) (x&((~x)+1))
22 inline void Update(int i,int x)
23 {
24     for(;i<=N;i+=lowbit(i)) tr[i]+=x;
25 }
26 inline int Query(int x)
27 {
28     int ret=0;
29     for(;x;x-=lowbit(x)) ret+=tr[x];
30     return ret;
31 }
32
33 int main()
34 {
35     for(;scanf("%d",&n)&&n;ans=0)
36     {
37         memset(tr,0,sizeof(tr));
38         for(int i=1;i<=n;i++)
39             scanf("%d",&a[i].num),a[i].mark=i;
40         sort(a+1,a+n+1,cmp);
41         for(int i=1;i<=n;Update(a[i++].mark,1))
42             ans+=(long long)Query(a[i].mark);
43         printf("%lld\n",ans);
44     }
45     return 0;
46 }