Ultra-QuickSort（离散化+树状数组求逆序数）

Link：http://poj.org/problem?id=2299

Ultra-QuickSort Time Limit: 7000MS   Memory Limit: 65536K Total Submissions: 44681   Accepted: 16243 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

[Submit]   [Go Back]   [Status]   [Discuss]

AC代码：

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<queue>
using namespace std;
struct node{
	int ord;
	int v;
}a[500001];
int n,c[500001],aa[500001];
int lowbit(int p)
{
	return p&(-p);
}
int sum(int p)
{
	int s=0;
	while(p>0)
	{
		s+=c[p];
		p-=lowbit(p);
	}
	return s;
}
void update(int p,int n,int k)
{
	while(p<=n)
	{
		c[p]+=k;
		p+=lowbit(p);
	}
}
bool cmp(node a,node b)
{
	return a.v<b.v;
}
int main()
{
	__int64 ans;
	while(scanf("%d",&n)&&n)
	{
		memset(aa,0,sizeof(aa));
		memset(c,0,sizeof(c));
		for(int i=1;i<=n;i++)
		{
			scanf("%d",&a[i].v);
			a[i].ord=i;
		}
		sort(a+1,a+n+1,cmp);
		for(int i=1;i<=n;i++)
		{
			aa[a[i].ord]=i;
		}
		ans=0;
		for(int i=1;i<=n;i++)
		{
			update(aa[i],n,1);
			ans+=(i-sum(aa[i]));
		}
		printf("%I64d\n",ans);
	}
	return 0;
}