#include <bits/stdc++.h>

using namespace std;

const int maxn = 2e5+5;

// 思路：建两个线段树，一个以L为1R为0  另一个相反
// 这样问题分解为求两个线段树的最长连续1的长度
struct Node{
	int Lsum;// 左连续1个数
	int Rsum;// 右连续1个数
	int val; // 区间最大连续1个数
}wL[maxn*4],wR[maxn*4];


void pushupL(int u,int L,int R)
{
	Node left = wL[2*u],right = wL[2*u+1];
	int M = L + R >> 1;
	wL[u].Lsum = (left.val==(M-L+1)) ? left.val + right.Lsum : left.Lsum;
	wL[u].Rsum = (right.val==(R-M)) ? right.val + left.Rsum : right.Rsum;
	wL[u].val = max(max(left.val,right.val),left.Rsum+right.Lsum);
}

void updateL(int u,int L,int R,int k)
{
	if(L==R)
	{
		wL[u].val = !wL[u].val;
		wL[u].Lsum = !wL[u].Lsum;
		wL[u].Rsum = !wL[u].Rsum;
		return;
	}
	int M = L + R >> 1;
	if(k<=M)
	{
		updateL(2*u,L,M,k);
	}else {
		updateL(2*u+1,M+1,R,k);
	}
	pushupL(u,L,R);
}


bool InRange(int L,int R,int l,int r)
{
	return (L>=l)&&(R<=r);
}

bool OutofRange(int L,int R,int l,int r)
{
	return (L>r) || (R<l);
}

Node queryL(int u,int L,int R,int l,int r) {
	if(InRange(L,R,l,r))
	{
		return wL[u];
	}
	if(OutofRange(L,R,l,r))
	{
		return Node{0,0,0};
	}
	int M = L + R >> 1;
	Node left = queryL(2*u,L,M,l,r);
	Node right = queryL(2*u+1,M+1,R,l,r);
	Node ans;
	ans.Lsum = (left.val==(M-L+1)) ? left.val + right.Lsum : left.Lsum;
	ans.Rsum = (right.val==(R-M)) ? right.val + left.Rsum : right.Rsum;
	ans.val = max(max(left.val,right.val),left.Rsum+right.Lsum);
	return ans;
}

void pushupR(int u,int L,int R)
{
	Node left = wR[2*u],right = wR[2*u+1];
	int M = L + R >> 1;
	wR[u].Lsum = (left.val==(M-L+1)) ? left.val + right.Lsum : left.Lsum;
	wR[u].Rsum = (right.val==(R-M)) ? right.val + left.Rsum : right.Rsum;
	wR[u].val = max(max(left.val,right.val),left.Rsum+right.Lsum);
}

void updateR(int u,int L,int R,int k)
{
	if(L==R)
	{
		wR[u].val = !wR[u].val;
		wR[u].Lsum = !wR[u].Lsum;
		wR[u].Rsum = !wR[u].Rsum;
		return;
	}
	int M = L + R >> 1;
	if(k<=M)
	{
		updateR(2*u,L,M,k);
	}else {
		updateR(2*u+1,M+1,R,k);
	}
	pushupR(u,L,R);
}

Node queryR(int u,int L,int R,int l,int r) {
	if(InRange(L,R,l,r))
	{
		return wR[u];
	}
	if(OutofRange(L,R,l,r))
	{
		return Node{0,0,0};
	}
	int M = L + R >> 1;
	Node left = queryR(2*u,L,M,l,r);
	Node right = queryR(2*u+1,M+1,R,l,r);
	Node ans;
	ans.Lsum = (left.val==(M-L+1)) ? left.val + right.Lsum : left.Lsum;
	ans.Rsum = (right.val==(R-M)) ? right.val + left.Rsum : right.Rsum;
	ans.val = max(max(left.val,right.val),left.Rsum+right.Lsum);
	return ans;
}

int main()
{
	int n,q;
	cin >> n >> q;
	for(int i = 1; i <= n; i++)
	{
		updateL(1,1,n,i); // 初始全为L
	}
	for(int i = 1; i <= q; i++)
	{
		int x;
		cin >> x;
		updateL(1,1,n,x);
		updateR(1,1,n,x);
		cout << max(queryL(1,1,n,1,n).val,queryR(1,1,n,1,n).val) << endl;
	}
}

对于这个代码主要解决的是求最长连续0或1的串 初始时L为全1，R为全0