Algorithm study notes: Cantor expansion & inverse Cantor expansion

1 Overview

Cantor Expansion & Inverse Cantor Expansion are two commonly used algorithms in full permutation problems.

Cantor Expansion: Knowing a $n$ order full array$a$ , find out which full permutation is this (sorted in lexicographic order).

Inverse Cantor Expansion: Knowing a $n$ order full array$a$ ranking obtained in this whole arrangement.

It can be found that these two operations are simply twin brothers.

2. Realize

2.1 Cantor expansion

example

Cantor’s expansion formula: $a_1\times (n-1)!+a_2\times (n-2)!+...+a_{n-1}\times 1!+a_n\times 0!+1$ . Conventions in this blog:$0!=0$。

In the above formula, $a_i$Means better than $a_i$Small and not in iiThe number of numbers in front of the$i$ position.

So we have to 5 2 3 1 4, for example, expanded simulation step Cantor.

1. Than the 5number of small four, which 0at the front, so $a_1=4$ . This4number can be generated$4\times 4!$ Full arrangement.
2. Than the 2number of a small, which has 0at the front, so $a_2=1$ . Taking into account of1the number has been determined, the remaining number capable of producing the$1\times 3!$ Full arrangement.
3. (Some words are omitted here)
4. The final result is $4\times 4!+1\times 3!+1\times 2!+0\times 1!+0\times 0!+1=105$

Why do you do this?

Such as the first one, we can find: For 1 2 3 4these four numbers, we need to calculate 5not the whole arrangement first. According to the knowledge of permutation and combination, the result is $4\times 4\times 3\times 2\times 1=4\times 4!$ . The rest is the same.

What about the code? In fact, it is easier to write.

Calculation $a_i$ We can use the weighted line segment tree, set, etc. to solve it.

Of course I used FHQ Treap, but it got stuck.

Never learned FHQ Treap students can take a look at this blog , there are very detailed explanation. Of course, you can also use other implementations.

Code:

#include <bits/stdc++.h>
using namespace std;

typedef long long LL;
const int MAXN = 1e6 + 10, P = 998244353;
int n, a[MAXN], cnt, root;
LL ans, f[MAXN];

struct node
{
    
    
	int l, r, size, val, key;
}tree[MAXN];

int read()
{
    
    
	int sum = 0, fh = 1; char ch = getchar();
	while (ch < '0' || ch > '9') {
    
    if (ch == '-') fh = -1; ch = getchar();}
	while (ch >= '0' && ch <= '9') {
    
    sum = (sum << 3) + (sum << 1) + (ch ^ 48); ch = getchar();}
	return sum * fh;
}

int Make_Node(int val)
{
    
    
	int now = ++cnt;
	tree[now].size = 1;
	tree[now].val = val;
	return now;
}

void update(int x) {
    
    tree[x].size = tree[tree[x].l].size + tree[tree[x].r].size + 1;}

void split(int now, int val, int &x, int &y)
{
    
    
	if (now == 0) x = y = 0;
	else
	{
    
    
		if (tree[now].val <= val)
		{
    
    
			x = now;
			split(tree[now].r, val, tree[now].r, y);
		}
		else
		{
    
    
			y = now;
			split(tree[now].l, val, x, tree[now].l);
		}
		update(now);
	}
}

int merge(int x, int y)
{
    
    
	if (!x || !y) return x + y;
	if (rand() & 1)
	{
    
    
		tree[x].r = merge(tree[x].r, y);
		update(x); return x;
	}
	else
	{
    
    
		tree[y].l = merge(x, tree[y].l);
		update(y); return y;
	}
}

void Insert(int val)
{
    
    
	int x, y; split(root, val - 1, x, y);
	root = merge(merge(x, Make_Node(val)), y);
}

int Find(int val)
{
    
    
	int x, y, ans;
	split(root, val - 1, x, y);
	ans = tree[x].size;
	root = merge(x, y);
	return ans;
}

int main()
{
    
    
	srand(time(0));
	n = read(); f[0] = 1; cnt = root = 0;
	for (int i = 1; i <= n; ++i) a[i] = read();
	for (int i = 1; i <= n; ++i) f[i] = f[i - 1] * i % P;
	f[0] = 0;
	for (int i = 1; i <= n; ++i)
	{
    
    
		Insert(a[i]); ans = (ans + ((LL)a[i] - 1 - Find(a[i])) * f[n - i]) % P;
//		printf("%d/%d/%d\n", f[n - i], Find(a[i]), ans);//调试用
	}
	printf("%lld\n", ans + 1);
	return 0;
}

2.2 Inverse Cantor expansion

Because the author is too weak, he did not learn.