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

Sample Output

6
0

这道题可以用线段树做，也可以用树状数组做，这道题树状数组快了一倍啊。。题目求的就是整个排序的逆序数，可以先离散化，然后每插入一个数，就判断前面有几个数（即比它小的数的个数）sum[i]，然后在这个数前且比这个数大的数的个数为i-sum[i],把它们都加起来就行了。逆序数的定义：排在pi前面并且比pi大的元素的个数称为元素pi的逆序数。

线段树代码：

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<map>
#include<queue>
#include<stack>
#include<string>
#include<algorithm>
using namespace std;
#define maxn 500010
__int64 sum;
struct node{
	int id,num,num1;
}a[maxn];

struct edge{
	int l,r,num;
}b[4*maxn];

bool cmp(node a,node b){
	return a.num<b.num;
}

bool cmp1(node a,node b){
	return a.id<b.id;
}

void build(int l,int r,int i)
{
	int mid;
	b[i].l=l;b[i].r=r;b[i].num=0;
	if(l==r)return;
	mid=(l+r)/2;
	build(l,mid,i*2);
	build(mid+1,r,i*2+1);
}

void question(int id,int i)
{
	int mid;
	if(b[i].l==b[i].r){
		b[i].num=1;return;
	}
	mid=(b[i].l+b[i].r)/2;
	if(id>mid)question(id,i*2+1);
	else {
		sum+=b[i*2+1].num;question(id,i*2);
	}
	b[i].num=b[i*2].num+b[i*2+1].num;
}

int main()
{
	int n,m,i,j,c;
	while(scanf("%d",&n)!=EOF && n!=0)
	{
		build(1,maxn,1);
		for(i=1;i<=n;i++){
			scanf("%d",&a[i].num);
			a[i].id=i;
		}
		sort(a+1,a+1+n,cmp);
		for(i=1;i<=n;i++){
			a[i].num1=i;
		}
		sort(a+1,a+1+n,cmp1);
		sum=0;
		for(i=1;i<=n;i++){
			question(a[i].num1,1);
		}
		printf("%I64d
",sum);
	}
	return 0;
}

树状数组代码：

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<map>
#include<queue>
#include<stack>
#include<string>
#include<algorithm>
using namespace std;
#define maxn 500005
struct node{
	int id,num;
}a[maxn];
int c[maxn];
bool cmp1(node a,node b){
	return a.num<b.num;
}
bool cmp2(node a,node b){
	return a.id<b.id;
}

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

void update(int pos,int num)
{
	while(pos<=maxn){
		c[pos]+=num;pos+=lowbit(pos);
	}
}

int sum(int pos)
{
	int num=0;
	while(pos>0){
		num+=c[pos];pos-=lowbit(pos);
	}
	return num;
}

int main()
{
	int n,m,i,j;
	__int64 num;
	while(scanf("%d",&n)!=EOF && n!=0)
	{
		for(i=1;i<=n;i++){
			scanf("%d",&a[i].num);
			a[i].id=i;
		}
		sort(a+1,a+1+n,cmp1);
		for(i=1;i<=n;i++) a[i].num=i;
		sort(a+1,a+1+n,cmp2);
		memset(c,0,sizeof(c));
		num=0;
		for(i=1;i<=n;i++){
			update(a[i].num,1);
			num+=i-sum(a[i].num);
		}
		printf("%I64d
",num);
	}
	return 0;
}