当前位置：网站首页>[code + 1] Yazid's freshman ball

[code + 1] Yazid's freshman ball

2021-08-08 00:30:56 【mrclr】

delivery

I took the original question in the competition , Gu Zige pushed out a very complex line segment tree , I didn't expect it to be a positive solution .

Our enumeration size is $x$ Number of numbers , So if in the interval $[L, R]$ in ,$x$ Meet the requirements of the topic , remember $num[i]$ by $1$ To $i$ in $x$ Number of occurrences of , So there are $\frac{num[R] - num[L - 1]}{R - L + 1} > \frac1{2}$, Put it in order $$2num[R] - R > 2num[L - 1] - (L - 1)$$
So we make $b[i] = 2num[i] - i$, It becomes to count $b[j] < b[i](j < i)$ I've got a lot of them .

Use whatever data structure to maintain , For size $x$ Statistics of the number of , It's about to reach $O(n^2logn)$ Complexity .

But every time from $1$ To $n$ Of course not , So we thought , Can you just traverse $x$ Position of appearance .

remember $i$ by $x$ The location of this occurrence ,$nxt[i]$ by $x$ The next place , So we have to find all the... Quickly $j$, Satisfy $b[j] < b[i], j < i, i \in [i, nxt[i] - 1]$.

about $(i, nxt[i] - 1]$ Medium $b[i]$, Yes $b[i] = b[i - 1] - 1$, So what you want to query is actually a continuous prefix and , namely $sum[b[i]], sum[b[i]-1], \cdots, sum[b[i] - (nxt[i] - i - 1)]$.

Because there are still changes , Therefore, we are inspired to use the segment tree to maintain the prefix and , That is, each node maintains the sum of all numbers less than or equal to the right endpoint .

Then the above inquiry is equivalent to the interval $[b[i] - (nxt[i] - i - 1) - 1, b[i] - 1]$ The query .

And modify , If you add a number $t$, Because it maintains prefixes and , To set the interval $[t, n]$ all $1$. But now we should put $[i, nxt[i] - 1]$ All contributions are added to the segment tree , About to $[b[i], n],[b[i] - 1, n], \cdots, [b[i] - (nxt[i] - i), n]$ Add it all $1$.

Think about it a little bit , Will find , In fact, it is the interval $[b[i] - (nxt[i] - i), b[i]]$ Add a first item as 1 A series of equal proportions , Then put the interval $[b[i] + 1, n]$ Add $nxt[i] - i$. Therefore, the segment tree should support interval addition and interval addition arithmetic sequence , And sum intervals .

Add equal ratio sequence , Convert to absolute subscript , Maintaining a one-time function .

#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<vector>
#include<queue>
#include<assert.h>
#include<ctime>
using namespace std;
//#define int long long
#define enter puts("") 
#define space putchar(' ')
#define Mem(a, x) memset(a, x, sizeof(a))
#define In inline
#define forE(i, x, y) for(int i = head[x], y; ~i && (y = e[i].to); i = e[i].nxt)
typedef long long ll;
typedef double db;
const int INF = 0x3f3f3f3f;
const db eps = 1e-8;
const int maxn = 5e5 + 5;
In ll read()
{
	ll ans = 0;
	char ch = getchar(), las = ' ';
	while(!isdigit(ch)) las = ch, ch = getchar();
	while(isdigit(ch)) ans = (ans << 1) + (ans << 3) + ch - '0', ch = getchar();
	if(las == '-') ans = -ans;
	return ans;
}
In void write(ll x)
{
	if(x < 0) x = -x, putchar('-');
	if(x >= 10) write(x / 10);
	putchar(x % 10 + '0');
}

int n, a[maxn];

int pos[maxn], nxt[maxn];
struct Tree
{
	int l, r;
	bool cl;
	ll lzi, lzy, sum;
	In void Clear()
	{
		cl = 1;
		lzi = lzy = sum = 0;
	}
}t[maxn << 3];
In void build(int L, int R, int now)
{
	t[now].l = L, t[now].r = R; 
	if(L == R) return (void)t[now].Clear();
	int mid = (L + R) >> 1;
	build(L, mid, now << 1), build(mid + 1, R, now << 1 | 1);
}
In void Change(int now, ll d1, ll d2)
{
	int len = t[now].r - t[now].l + 1;
	t[now].lzi += d1;
	t[now].sum += (d1 * (1LL * t[now].l + t[now].r) * len >> 1);
	t[now].lzy += d2;
	t[now].sum += d2 * len;
}
In void pushdown(int now)
{
	if(t[now].cl) t[now << 1].Clear(), t[now << 1 | 1].Clear(), t[now].cl = 0;
	if(t[now].lzi || t[now].lzy)
	{
		Change(now << 1, t[now].lzi, t[now].lzy);
		Change(now << 1 | 1, t[now].lzi, t[now].lzy);
		t[now].lzi = t[now].lzy = 0;
	}
}
In void update(int L, int R, int now, int d1, int d2)
{
	if(L > R) return;
	if(t[now].l == L && t[now].r == R)
	{
		Change(now, d1, d2);
		return;
	}
	pushdown(now);
	int mid = (t[now].l + t[now].r) >> 1;
	if(R <= mid) update(L, R, now << 1, d1, d2);
	else if(L > mid) update(L, R, now << 1 | 1, d1, d2);
	else update(L, mid, now << 1, d1, d2), update(mid + 1, R, now << 1 | 1, d1, d2);
	t[now].sum = t[now << 1].sum + t[now << 1 | 1].sum;
}
In ll query(int L, int R, int now)
{
	if(t[now].l == L && t[now].r == R) return t[now].sum;
	pushdown(now);
	int mid = (t[now].l + t[now].r) >> 1;
	if(R <= mid) return query(L, R, now << 1);
	else if(L > mid) return query(L, R, now << 1 | 1);
	else return query(L, mid, now << 1) + query(mid + 1, R, now << 1 | 1) ;
}

In ll solve(int p)
{
	nxt[0] = p;
	int x = 0, num = 0; ll ret = 0;
	while(x <= n)
	{
		if(x > 0) num++;
		int val = (num << 1) - x, tp = nxt[x] - x - 1;
		ret += query(val - tp - 1, val - 1, 1);
		update(val - tp, val, 1, 1, -(val - tp - 1));
		update(val + 1, n, 1, 0, tp + 1);
		x = nxt[x]; 
	}
	t[1].Clear();
	return ret;
}

int main()
{
	n = read(); int TYPE = read();
	for(int i = 1; i <= n; ++i) a[i] = read(), pos[i - 1] = n + 1;
	for(int i = n; i; --i) nxt[i] = pos[a[i]], pos[a[i]] = i;
	Mem(pos, 0);		// Repeated use  
	build(-n, n, 1);
	ll ans = 0;
	for(int i = 1; i <= n; ++i) if(!pos[a[i]]) ans += solve(i), pos[a[i]] = 1;
	write(ans), enter;
	return 0;
}

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版权声明
本文为[mrclr]所创，转载请带上原文链接，感谢
https://chowdera.com/2021/08/20210808003020469S.html

当前位置：网站首页>[code + 1] Yazid's freshman ball

[code + 1] Yazid's freshman ball

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