POJ——T 2299 Ultra-QuickSort

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

Time Limit: 7000MS		Memory Limit: 65536K
Total Submissions: 62894		Accepted: 23442

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

 #include <algorithm>
 #include <cstring>
 #include <cstdio>

 using namespace std;

 const int N();
 long long ans;
 int n,tr[N];
 struct Node
 {
     int num,mark;
 }a[N];
 bool cmp(Node a,Node b)
 {
     if(a.num==b.num)
         return a.mark>b.mark;
     return a.num>b.num;
 }

 #define lowbit(x) (x&((~x)+1))
 inline void Update(int i,int x)
 {
     for(;i<=N;i+=lowbit(i)) tr[i]+=x;
 }
 inline int Query(int x)
 {
     int ret=;
     for(;x;x-=lowbit(x)) ret+=tr[x];
     return ret;
 }

 int main()
 {
     for(;scanf("%d",&n)&&n;ans=)
     {
         memset(tr,,sizeof(tr));
         for(int i=;i<=n;i++)
             scanf("%d",&a[i].num),a[i].mark=i;
         sort(a+,a+n+,cmp);
         for(int i=;i<=n;Update(a[i++].mark,))
             ans+=(long long)Query(a[i].mark);
         printf("%lld\n",ans);
     }
     return ;
 }