POJ2299——Ultra-QuickSort

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

树状数组求逆序数....再加个离散化

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

using namespace std;

int tree[500010];
int val[500010];
int n;
struct node
{
	long long val;
	int index;
}arr[500010];

int cmp(node a,node b)
{
	return a.val < b.val;
}

inline int lowbit(int x)
{
	return x&(-x);
}

void add(int x,int val)
{
	for(int i=x;i<=n;i+=lowbit(i))
		tree[i]+=val;
}

long long sum(int x)
{
	long long ans=0;
	for(int i=x;i;i-=lowbit(i))
		ans+=tree[i];
	return ans;
}

int main()
{
	while(~scanf("%d",&n),n)
	{
		for(int i=1;i<=n;i++)
		{
			scanf("%d",&arr[i].val);
			arr[i].index=i;
		}
		sort(arr+1,arr+n+1,cmp);
		for(int i=1;i<=n;i++)
		{
			if(i==1)
				val[arr[i].index]=i;
			else
			{
				if(arr[i].val==arr[i-1].val)
					val[arr[i].index]=val[arr[i-1].index];
				else
					val[arr[i].index]=i;
			}
		}
			
		long long ans=0;
		memset(tree,0,sizeof(tree));
		for(int i=1;i<=n;i++)
		{
			add(val[i],1);
			ans+=i-sum(val[i]);
		}
		printf("%lld\n",ans);
	}
	return 0;
}