poj2299 Ultra-QuickSort

Ultra-QuickSort

Time Limit: 7000MS		Memory Limit: 65536K
Total Submissions: 43339		Accepted: 15798

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

自从高二以来好像只用线段树没有写过树状数组……随便拿个求逆序对的题练练手

#include<cstdio>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#include<queue>
#include<deque>
#include<set>
#include<map>
#include<ctime>
#define LL long long
#define inf 0x7ffffff
#define pa pair<int,int>
#define pi 3.1415926535897932384626433832795028841971
using namespace std;
inline LL read()
{
    LL x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
struct dat{
	int a,rnk;
}a[500010];
int d[500010];
int c[500010];
bool operator <(const dat &a,const dat &b)
{return a.a<b.a;}
int n,cnt;
LL ans;
inline int lowbit(int x){return x&(-x);}
inline int change(int x)
{
	for (int i=x;i<=n;i+=lowbit(i))
		c[i]++;
}
inline int query(int x)
{
	int sum=0;
	for (int i=x;i;i-=lowbit(i))
		sum+=c[i];
	return sum;
}
int main()
{
	while (scanf("%d",&n)&&n)
	{
		for (int i=1;i<=n;i++)
			a[i].a=read(),a[i].rnk=i;
		sort(a+1,a+n+1);
		cnt=ans=0;
		a[0].a=-1;
		for (int i=1;i<=n;i++)
		{
			if (a[i].a!=a[i-1].a)cnt++;
		  	d[a[i].rnk]=cnt;
		}
		memset(c,0,sizeof(c));
		for (int i=n;i>=1;i--)
		{
		  	ans+=query(d[i]);
		  	change(d[i]);
		}
		printf("%lld
",ans);
	}
}