#include<bits/stdc++.h>
using namespace std;
const int maxn=1e6+5;
inline char gc(){
    static char buf[1000000],*p1=buf,*p2=buf;
    return p1==p2&&(p2=(p1=buf)+fread(buf,1,1000000,stdin),p1==p2)?EOF:*p1++;
}
inline int read(){
	int res=0;
	char ch=gc();
	while(ch<'0'||ch>'9')
		ch=gc();
	while(ch>='0'&&ch<='9'){
		res=(res<<1)+(res<<3)+(ch^'0');
		ch=gc();
	}
	return res;
}
int n,a[maxn],k=1,ans,f;
struct node{
	int len,check,id;
	bool operator <(const node &t) const{
		if(len==t.len)
			return id>t.id;
		return len<t.len;
	}
};
vector<node> s;
priority_queue<node> q;
bool vis[maxn];
inline int dfs(int x){
	int l=0,r=s.size()-1;
	while(l<r){
		int mid=l+r>>1;
		int k=s[mid].id;
		if(k==x)
			return mid;
		if(k>x)
			r=mid-1;
		else
			l=mid+1;
	}
	return r;
}
int main(){
	n=read();
	for(int i=1;i<=n;++i){
		a[i]=read();
		if(i!=1&&a[i]!=a[i-1]){
			s.push_back(node({i-k,a[i-1],++f}));
			q.push(node({i-k,a[i-1],f}));
			k=i;
		}
		if(i==n){
			s.push_back(node({n-k+1,a[n],++f}));
			q.push(node({n-k+1,a[n],f}));
		}
	}
	while(!q.empty()){
		int x=q.top().id;
		q.pop();
		if(vis[x])
			continue;
		vis[x]=1;
		int k=dfs(x);
		++ans;
		if(k&&k!=s.size()-1){
			node A=s[k-1],B=s[k+1];
			if(A.check==B.check){
				A.len+=B.len;
				vis[B.id]=1;
				q.push(A);
				s[k-1]=A;
				s.erase(s.begin()+k+1);
			}
		}
		s.erase(s.begin()+k);
	}
	printf("%d\n",ans);
	return 0;
}

时间复杂度是 $O(n\log^2 n)$，但它为什么会 TLE 啊（常数？）？