description
analysis
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可以貪心還原出原序列,且是的排列;易知由是連續若干段的單調不遞減區間拼起來而成
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而且每一段區間內差值至多爲,大概像這樣
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對每一段區間來說,前一塊中的數都小於後一塊中的數;而且每一塊中數遞減填最優
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最後一段的數填越小也越優,考慮限制關係,從出現最後的位置向連邊,表示形成一棵樹
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由於前向星連邊越晚的越早遍歷,便可以確保越後的位置越小,後時間戳即爲
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最後再用一個樹狀數組求即可
code
#pragma GCC optimize("O3")
#pragma G++ optimize("O3")
#include<stdio.h>
#include<string.h>
#include<algorithm>
#define MAXN 100005
#define MAXM MAXN*2
#define ll long long
#define reg register ll
#define fo(i,a,b) for (reg i=a;i<=b;++i)
#define fd(i,a,b) for (reg i=a;i>=b;--i)
#define rep(i,a) for (reg i=las[a];i;i=nex[i])
using namespace std;
ll las[MAXM],nex[MAXM],tov[MAXM];
ll a[MAXN],dfn[MAXN],last[MAXN];
ll tr[MAXN];
ll n,tot,ans;
inline ll read()
{
ll x=0,f=1;char ch=getchar();
while (ch<'0' || '9'<ch){if (ch=='-')f=-1;ch=getchar();}
while ('0'<=ch && ch<='9')x=x*10+ch-'0',ch=getchar();
return x*f;
}
inline ll lowbit(ll x){return x&(-x);}
inline ll max(ll x,ll y){return x>y?x:y;}
inline ll min(ll x,ll y){return x<y?x:y;}
inline void dfs(ll x){dfn[x]=++tot;rep(i,x)dfs(tov[i]);}
inline void link(ll x,ll y){nex[++tot]=las[x],las[x]=tot,tov[tot]=y;}
inline void modify(ll x,ll y){while (x<=n)tr[x]=max(tr[x],y),x+=lowbit(x);}
inline ll query(ll x){ll y=0;while (x)y=max(y,tr[x]),x-=lowbit(x);return y;}
int main()
{
//freopen("alice.in","r",stdin);
//freopen("alice.out","w",stdout);
n=read();
fo(i,1,n)a[i]=read(),link(last[a[i]-1],i),last[a[i]]=i;
tot=-1,dfs(0);
fo(i,1,n)printf("%lld ",dfn[i]);
printf("\n");
fd(i,n,1)
{
ll tmp=query(dfn[i]-1)+1;
ans+=tmp,modify(dfn[i],tmp);
}
printf("%lld\n",ans);
return 0;
}