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(500000);
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=0;
    for(;x;x-=lowbit(x)) ret+=tr[x];
    return ret;
}

int main()
{
    for(;scanf("%d",&n)&&n;ans=0)
    {
        memset(tr,0,sizeof(tr));
        for(int i=1;i<=n;i++)
            scanf("%d",&a[i].num),a[i].mark=i;
        sort(a+1,a+n+1,cmp);
        for(int i=1;i<=n;Update(a[i++].mark,1))
            ans+=(long long)Query(a[i].mark);
        printf("%lld\n",ans);
    }
    return 0;
}