poj 2299 Ultra-QuickSort（求逆序对）

Ultra-QuickSort

Time Limit: 7000MS		Memory Limit: 65536K
Total Submissions: 52778		Accepted: 19348

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

题意：给一组数，每次只能交换相邻两个数的位置，问最少交换多少次，使得这个序列从小到大排序；
题解：求逆序数的对数

线段树：

#include<stdio.h>
#include<string.h>
#include<algorithm>
#define LL long long
#define MAX 500500
using namespace std;
int n,m;
struct node
{
	int val,pos;
}num[MAX];
int sum[MAX<<2];
void pushup(int o)
{
	sum[o]=sum[o<<1]+sum[o<<1|1];
}
void gettree(int o,int l,int r)
{
	sum[o]=0;
	if(l==r)
		return ;
	int mid=(l+r)>>1;
	gettree(o<<1,l,mid);
	gettree(o<<1|1,mid+1,r);
	pushup(o);
}

void update(int o,int l,int r,int pos)
{
	if(l==r)
	{
		sum[o]+=1;
		return ;
	}
	int mid=(l+r)>>1;
	if(pos<=mid) update(o<<1,l,mid,pos);
	else update(o<<1|1,mid+1,r,pos);
	pushup(o);
}
int find(int o,int l,int r,int L,int R)
{
	if(L<=l&&R>=r)
	return sum[o];
	int mid=(l+r)>>1;
	int ans=0;
	if(L<=mid) ans+=find(o<<1,l,mid,L,R);
	if(R>mid) ans+=find(o<<1|1,mid+1,r,L,R);
	return ans;
}
bool cmp(node a,node b)
{
	if(a.val!=b.val) return a.val<b.val;
	else return a.pos<b.pos;
}
int main()
{
	int j,i,t,k;
	while(scanf("%d",&n),n)
	{
		gettree(1,1,n);
		for(i=1;i<=n;i++)
		{
			scanf("%d",&num[i].val);
			num[i].pos=i;
		}
		sort(num+1,num+1+n,cmp);
		LL ant=0;
		for(i=1;i<=n;i++)
		{
			update(1,1,n,num[i].pos);
			ant+=i-find(1,1,n,1,num[i].pos);
		}
		printf("%lld\n",ant);
	}
	return 0;
}

　　树状数组：

#include<stdio.h>
#include<string.h>
#include<algorithm>
#define LL long long
#define MAX 500500
using namespace std;
int n,m;
int c[MAX<<1];
struct node
{
	int val,pos;
}num[MAX];
bool cmp(node a,node b)
{
	if(a.val!=b.val)
	    return a.val<b.val;
	else
	    return a.pos<b.pos;
}
void update(int x)
{
	while(x<=n)
	{
		c[x]+=1;
		x+=x&-x;
	}
}
LL sum(int x)
{
	LL ans=0;
	while(x>=1)
	{
		ans+=c[x];
		x-=x&-x;
	}
	return ans;
}
int main()
{
	int j,i,t,k;
	while(scanf("%d",&n),n)
	{
		for(i=1;i<=n;i++)
		{
			scanf("%d",&num[i].val);
			num[i].pos=i;
		}
		sort(num+1,num+n+1,cmp);
		memset(c,0,sizeof(c));
		LL ant=0;
		for(i=1;i<=n;i++)
		{
			update(num[i].pos);
			ant+=i-sum(num[i].pos);
		}
		printf("%lld\n",ant);
	}
	return 0;
}