Bicolored RBS CodeForces

建树, 然后高度最大值的最小值显然为$lceil frac{dep}{2} ceil$, 将$>frac{dep}{2}$的全部分出去即可.

#include <sstream>
#include <algorithm>
#include <cstdio>
#include <math.h>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <string.h>
#include <bitset>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '
'
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<' ';hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, P2 = 998244353, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;}
//head



const int N = 1e6+10;
int n, tot, now, mx;
char s[N], Ans[N];
vector<int> g[N];
int L[N], R[N], dep[N];

void dfs(int x) {
	while (s[now]=='(') {
		L[tot+1]=now;
		++now,g[x].pb(++tot),dfs(tot);
	}
	R[x]=now++;
}
void dfs(int x, int d) { 
	mx=max(mx,dep[x]=d);
	for (int y:g[x]) dfs(y,d+1);
}
int main() {
	scanf("%d%s", &n, s+1);
	now=tot=1,dfs(1),dfs(1,0);
	memset(Ans,'0',sizeof Ans);
	REP(i,1,tot) if (dep[i]>mx/2) Ans[L[i]]=Ans[R[i]]='1';	
	Ans[n+1]=0;
	puts(Ans+1);
}