Minimum Inversion Number--线段树（单点更新）

题目    HDU--1394

The inversion number of a given number sequence a1, a2, ..., an is the number of pairs (ai, aj) that satisfy i < j and ai > aj.

For a given sequence of numbers a1, a2, ..., an, if we move the first m >= 0 numbers to the end of the seqence, we will obtain another sequence. There are totally n such sequences as the following:

a1, a2, ..., an-1, an (where m = 0 - the initial seqence)
a2, a3, ..., an, a1 (where m = 1)
a3, a4, ..., an, a1, a2 (where m = 2)
...
an, a1, a2, ..., an-1 (where m = n-1)

You are asked to write a program to find the minimum inversion number out of the above sequences.

Input

The input consists of a number of test cases. Each case consists of two lines: the first line contains a positive integer n (n <= 5000); the next line contains a permutation of the n integers from 0 to n-1.

Output

For each case, output the minimum inversion number on a single line.

Sample Input

10
1 3 6 9 0 8 5 7 4 2

Sample Output

16

题意

多组输入。给定一个数字n，然后第二行输入n个在0--n-1之间的数。在给定的序列中，每次把第一个数移到最后一个位置，求所有序列的逆序数的最小值。

分析

利用线段树求a[i]+1到n-1之间的大于a[i]的个数，统计完初始序列的逆序对数之后，就开始移动。每次把a[i]移动到最后一个位置之前，a[i]后面比a[i]大的有n-a[i]-1个，a[i]后面比a[i]小的有a[i]个，移动后，大于a[i]的全部成了逆序，小于a[i]的全部成了顺序，所以，在原来逆序数的基础上加上n-a[i]-1，再减去a[i]，最后跟原来比较。

代码

#include<iostream>
#include<cstdio>
#include<string.h>
using namespace std;
const int N=5005;
int a[N];
struct node{
	int left,right;
	int sum;
}tree[N<<2];
void pushup(int m)
{
	tree[m].sum=tree[m<<1].sum+tree[m<<1|1].sum;
}
void build(int m,int l,int r)
{
	tree[m].sum=0;
	tree[m].left=l;
	tree[m].right=r;
	if(l==r)
		return ;
	int mid=(l+r)>>1;
	build(m<<1,l,mid);
	build(m<<1|1,mid+1,r);
}
void updata(int m,int id)
{
	if(tree[m].left==id&&tree[m].right==id)
	{
		tree[m].sum=1;//出现过，标为1
		return ;
	}
	int mid=(tree[m].left+tree[m].right)>>1;
	if(id<=mid)
		updata(m<<1,id);
	else
		updata(m<<1|1,id);
	pushup(m);
}
int query(int m,int l,int r)
{
	if(tree[m].left>=l&&tree[m].right<=r)//
	{
		return tree[m].sum;
	}
	int mid=(tree[m].left+tree[m].right)>>1;
	int sum1=0,sum2=0;
	if(l<=mid)
		sum1=query(m<<1,l,r);
	if(r>mid)
		sum2=query(m<<1|1,l,r);
	return sum1+sum2;
}
int main()
{
	int n,sum;
	while(~scanf("%d",&n))
	{
		if(n==0)
			break;
		build(1,0,n-1);
		sum=0;
		for(int i=0;i<n;i++)
		{
			scanf("%d",&a[i]);
			sum += query(1,a[i]+1,n-1);//比a[i]大的数肯定在a[i]+1到n-1之间 
			updata(1,a[i]);
		}
		int ans=sum;
		for(int i=0;i<n;i++)
		{
			sum=sum+n-2*a[i]-1;//把a[i]移到最后面，a[i]后面大于a[i]的数有n-a[i]-1个，
			ans=min(ans,sum);//a[i]后面比a[i]小的有a[i]个，要加上大的，减去小的 
		}
		printf("%d\n",ans);
	}
	return 0;
}

标签：...,Inversion,int,Number,number,a1,Minimum,where,a2
来源： https://blog.csdn.net/Napom/article/details/89675337