HDU 1394 (树状数组 & 线段树两种做法) Minimum Inversion Number

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我是靠谱客的博主有魅力皮带，最近开发中收集的这篇文章主要介绍HDU 1394 (树状数组 & 线段树两种做法) Minimum Inversion Number，觉得挺不错的，现在分享给大家，希望可以做个参考。

概述

Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 16165 Accepted Submission(s): 9836

Problem Description

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

题目意思：求逆序数，然后将第一个数放到最后，求出逆序数，再将新的数组的第一个数放到最后，在求出新的逆序数，最后求这些逆序数中的最小的一个

首先用了线段树写的，首先将数组插入到线段树种，然后对从这个数到最后进行查询，就可以找出在插入的这个数之前的大于这个数的个数，这道题是按照找前面的比他大的数来累加的，和从后面找比他小的个数的和是一样的

#include <iostream>
#include <cstring>
#include <cstdio>
#define Max 5005
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1
using namespace std;

int segTree[Max<<2];//存的是树的下面的最小的值
int ary[Max];//存的是输入的数字
int vis[Max<<2];
//
inline max(int a,int b)
{
	return a>b?a:b;
}

inline min(int a,int b)
{
	return a>b?b:a;
}

void push_up(int rt)
{
	segTree[rt] = segTree[rt<<1] + segTree[rt<<1|1];
}

void build(int l,int r,int rt)//构建线段树，并求出每个节点的下面分支的最小值
{
	segTree[rt] = 0;

	if(l == r)
		return ;

	int m = (l+r)>>1;
	build(ls);
	build(rs);
	push_up(rt);
}

void update(int site,int l,int r,int rt)//单点更新
{
	if(l == r)
	{
		segTree[rt]++;
		return ;
	}
	
	int m = (l + r) >> 1;
	
	if(site <= m)
		update(site,ls);
	else
		update(site,rs);
	
	push_up(rt);
}

int query(int ll,int rr,int l,int r,int rt)
{

	if(l >= ll && r <= rr)
		return segTree[rt];

	int m = (l + r) >> 1;
	int ans =0 ;

	if(ll <= m)
		ans += query(ll,rr,ls);
	if(rr > m)
		ans += query(ll,rr,rs);

	return ans;
}

int main()
{
    int n,m;
	int i,j,k;
	int x[Max];

    while(cin>>n)
	{
		build(0,n-1,1);
		int sum = 0;

		for(i=0;i<n;i++)
		{
			scanf("%d",&x[i]);
			sum += query(x[i],n-1,0,n-1,1);
			update(x[i],0,n-1,1);
		}

		int res = sum;
		for(i=0;i<n;i++)
		{
			sum += n-x[i] - x[i] - 1;
			res = min(sum,res);
		}

		cout<<res<<endl;
    }  
    return 0; 
}

然后试着用树状数组也写了一下，原理和线段树是相同的，只不过树状数组找比他前面的大的数，是先找比他小的数，然后再拿总的数减去比他小的数就可以求出比他大的数了

#include <iostream>
#include <cstring>
#include <cstdio>
#define Max 5005
using namespace std;

int c[Max],nn;

inline max(int a,int b)
{
	return a>b?a:b;
}

inline min(int a,int b)
{
	return a>b?b:a;
}

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

void add(int k)
{
	while(k <= nn)
	{
		c[k]++;
		k += lowbit(k);
	}
}

int sum(int k)
{
	int s = 0;
	while(k)
	{
		s += c[k];
		k -= lowbit(k);
	}

	return s;
}

int main()
{
    int n,m;
	int i,j,k;
	int x[Max];

    while(cin>>nn)
	{
		m=0;
		memset(c,0,sizeof(c));

		for(i=0;i<nn;i++)
		{
			scanf("%d",&x[i]);
			m += (sum(nn) - sum(x[i]+1));
			add(x[i]+1);
		}

		int res = m;
		for(i=0;i<nn;i++)
        {
			m += nn - x[i] - x[i] -1;
			res = min(res,m);
		}

		cout<<res<<endl;
    }  
    return 0; 
}

最终发现树状数组还是效率更高一点点