用线段树解决逆序数问题(1394)

最新推荐文章于 2024-10-14 09:00:00 发布

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最新推荐文章于 2024-10-14 09:00:00 发布

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分类专栏：线段树

本文链接：https://blog.youkuaiyun.com/qq_33765907/article/details/50971715

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本文介绍了一种算法，用于解决给定序列通过不同轮转方式得到的所有序列中逆序数最小的问题。通过构建特定的数据结构，实现了快速计算逆序数及轮转后的贡献变化。

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Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 16195 Accepted Submission(s): 9851

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

注意这里的build函数不需要额外的操作，只需要建好树即可，下面其实就是很朴素的区间和问题，求出逆序数之后，还需要一个一个地把第一个数放到最后去，求解这样做产生的贡献，最后求得最小值。

/*------------------Header Files------------------*/
#include <iostream>
#include <cstring>
#include <string>
#include <cstdio>
#include <algorithm>
#include <cstdlib>
#include <ctype.h>
#include <cmath>
#include <stack>
#include <queue>
#include <deque>
#include <map>
#include <vector>
#include <limits.h>
using namespace std;
/*------------------Definitions-------------------*/
#define LL long long
#define PI acos(-1.0)
#define INF 0x3F3F3F3F
#define MOD 1E9+7
#define MAX 500050
#define lson rt<<1,l,m
#define rson rt<<1|1,m+1,r
/*---------------------Work-----------------------*/
int n,tree[5050<<2],num[5050];
void pushup(int rt)
{
	tree[rt]=tree[rt<<1]+tree[rt<<1|1];
}
void build(int rt,int l,int r) //初始化操作，此题的建树没有多余操作
{
	tree[rt]=0;
	if(l==r) return ;
	int m=(l+r)>>1;
	build(lson);
	build(rson);
}
void update(int rt,int l,int r,int p)
{
	if(l==r)
	{
		tree[rt]++;
		return ;
	}
	int m=(l+r)>>1;
	if(p<=m) update(lson,p);
	else update(rson,p);
	pushup(rt);
}
int query(int rt,int l,int r,int L,int R)
{
	if(L<=l&&R>=r) return tree[rt];
	int m=(l+r)>>1;
	int sum=0;
	if(L<=m) sum+=query(lson,L,R);
	if(R>m) sum+=query(rson,L,R);
	return sum;
}
void work()
{
	while(scanf("%d",&n)==1)
	{
		build(1,0,n-1); //区间只能0到n-1，源于题意，根结点不能从0开始
		int sum=0;
		for(int i=0;i<n;i++)
		{
			scanf("%d",&num[i]);
			sum+=query(1,0,n-1,num[i],n-1);
			update(1,0,n-1,num[i]);
		}
		//开始转换求最小值
		int ans=sum; //sum记录目前的逆序数值
		for(int i=0;i<n;i++)
		{
			sum=sum-num[i]+n-1-num[i];
			ans=min(ans,sum);
		}
		printf("%d\n",ans);
	}
}
/*------------------Main Function------------------*/
int main()
{
	//freopen("test.txt","r",stdin);
	//freopen("cowtour.out","w",stdout);
	//freopen("cowtour.in","r",stdin);
	work();
	return 0;
}