Luo Gu [P1903] [] BZOJ2120 number of colors / maintenance repair queue with Mo [team]

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Subject to the effect:

Topic links:
Luo Gu: https://www.luogu.org/problemnew/show/P1903
BZOJ: https://www.lydsy.com/JudgeOnline/problem.php?id=2120

$n$ only brush, maintain the following:

1. $Q\ L\ R$ representative asked from you $L$ brushes to the second $R$ brushes brush There are several different colors.
2. $R\ P\ Col$ the first $P$ brushes with a color $Col$。

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Ideas:

Beginning to see this exciting topic knocked a block, knocking the last to find $col \ leq 10 ^ 6$ rather than $10^5$ . . .
Correct answer is with Mo repair team.
If there is no modification operations, then this is the team of Mo board.
Just add modify it.

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In fact, the team Mo modifications are violent.
We call each modified record, and then modify the number before each inquiry needed to be recorded. If you are asked $i$ ratio inquiry $i-1$ a few more changes, then put these changes changing for the better, if less then change it back.
In fact, really is violence.

void work(int x,int i)  //要做第x次修改，这次是询问i
{
	if (chg[x].x>=ask[i].l&&chg[x].x<=ask[i].r)  //在区间内修改才有贡献
	{
		cnt[a[chg[x].x]]--;
		if (!cnt[a[chg[x].x]]) sum--;
		if (!cnt[chg[x].val]) sum++;
		cnt[chg[x].val]++; 
	}
	swap(a[chg[x].x],chg[x].val);
	//交换之后，如果下次要改回来时就相当于把现在的值储存到原数组里
}

Mo repair team with the most average complexity is about to take $Block=\sqrt[3]{n}$. It proved to be impossible to prove the XD
do not know why I knocked the thief slow mo running team. do not add $Ofast,inline$ , then run $4s +$ . Please Gangster treatise.

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Code:

%:pragma GCC optimize("Ofast")
%:pragma GCC optimize("inline")
#include <cmath>
#include <cstdio>
#include <string>
#include <algorithm>
#define rr register
using namespace std;

const int N=50010,M=1000010;
int n,m,s,Q,T,l,r,sum,k,a[N],ans[N],cnt[M],pos[N];
char ch;

int read()
{
	int d=0;
	char ch=getchar();
	while (!isdigit(ch)) ch=getchar();
	while (isdigit(ch))
		d=(d<<3)+(d<<1)+ch-48,ch=getchar();
	return d;
}

void write(int x)
{
	if (x>9) write(x/10);
	putchar(x%10+48);
}

struct Ask
{
	int l,r,id,k;
}ask[N];

struct Change
{
	int x,val;
}chg[N];

bool cmp(Ask x,Ask y)
{
	if (pos[x.l]<pos[y.l]) return 1;
	if (pos[x.l]>pos[y.l]) return 0;
	return x.r<y.r;
}

void del(int x)
{
	cnt[a[x]]--;
	if (!cnt[a[x]]) sum--;
}

void add(int x)
{
	if (!cnt[a[x]]) sum++;
	cnt[a[x]]++;
}

void work(int x,int i)
{
	if (chg[x].x>=ask[i].l&&chg[x].x<=ask[i].r)
	{
		cnt[a[chg[x].x]]--;
		if (!cnt[a[chg[x].x]]) sum--;
		if (!cnt[chg[x].val]) sum++;
		cnt[chg[x].val]++;
	}
	swap(a[chg[x].x],chg[x].val);
}

int main()
{
	n=read(); Q=read();
	T=pow((double)n,(1.0/3.0));
	for (rr int i=1;i<=n;i++)
	{
		a[i]=read();
		pos[i]=(i-1)/T+1;
	}
	for (rr int i=1;i<=Q;i++)
	{
		while (ch=getchar()) if (ch=='Q'||ch=='R') break;
		if (ch=='Q')
		{
			m++;
			ask[m].l=read(); ask[m].r=read();
			ask[m].id=m; ask[m].k=s;
		}
		else
		{
			s++;
			chg[s].x=read(); chg[s].val=read();
		}
	}
	sort(ask+1,ask+1+m,cmp);
	l=1;
	for (rr int i=1;i<=m;i++)
	{
		for (;l<ask[i].l;l++) del(l);
		for (;l>ask[i].l;l--) add(l-1);
		for (;r<ask[i].r;r++) add(r+1);
		for (;r>ask[i].r;r--) del(r);
		for (;k<ask[i].k;k++) work(k+1,i);
		for (;k>ask[i].k;k--) work(k,i);
		ans[ask[i].id]=sum;
	}
	for (rr int i=1;i<=m;i++)
		write(ans[i]),putchar(10);
	return 0;
}