Bicolored RBS CodeForces - 1167D (括号)
发布时间:2020-12-14 04:47:01 所属栏目:大数据 来源:网络整理
导读:建树,然后高度最大值的最小值显然为$lceil frac{dep}{2}rceil$,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
建树,然后高度最大值的最小值显然为$lceil frac{dep}{2}rceil$,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,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 ‘n‘ #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,tot) if (dep[i]>mx/2) Ans[L[i]]=Ans[R[i]]=‘1‘; Ans[n+1]=0; puts(Ans+1); } (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |