计蒜客-LIS

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题意

从长度为 $n(\le 5e5)$ 的数组中等概率的取出一个最长上升子序列, 求每个数被选中的概率

题解

$bg[i]$ 是以 $a[i]$ 为起点的 $LIS$ 的长度, $g[i]$ 是以 $a[i]$ 为起点的 $LIS$ 的数量

$en[i]$ 是以 $a[i]$ 为终点的 $LIS$ 的长度, $h[i]$ 是以 $a[i]$ 为终点的 $LIS$ 的数量

$\text{len_a}[i]$ 是以 $i$ 为起点或终点的 $LIS$ 的长度, $\text{cnt_a}[i]$ 是以 $i$ 为起点或终点的 $LIS$ 的数量

$\text{len_h}[i],\text{cnt_h}[i] $ 是树状数组的辅助数组

代码

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#include <bits/stdc++.h>
using namespace std;
#define forl(i, l, r) for (int i = l; i <= r; i++)
#define forr(i, r, l) for (int i = r; i >= l; i--)
#define for1(i, n) for (int i = 1; i <= n; i++)
#define fro1(i, n) for (int i = 1; i <= n; i++)
#define for0(i, n) for (int i = 0; i < n; i++)
#define fro0(i, n) for (int i = 0; i < n; i++)
#define meminf(a) memset(a, inf, sizeof(a))
#define mem_1(a) memset(a, -1, sizeof(a))
#define mem0(a) memset(a, 0, sizeof(a))
#define memcp(a,b) memcpy(a,b,sizeof(b))
#define oper(type) bool operator <(const type y)const
#define mp make_pair
#define pu_b push_back
#define pu_f push_front
#define po_b pop_back
#define po_f pop_front
#define fi first
#define se second
#define whiel while
#define retrun return
typedef pair<long long, long long> pll;
typedef vector<long long> vll;
typedef pair<int, int> pii;
typedef unsigned long long ull;
typedef vector<int> vii;
typedef long double db;
typedef long long ll;
typedef int itn;
int in(int &a,int &b,int &c,int &d){return scanf("%d%d%d%d",&a,&b,&c,&d);}
int in(int &a,int &b,int &c){return scanf("%d%d%d",&a,&b,&c);}
int in(int &a,int &b){return scanf("%d%d",&a,&b);}
int in(ll &a){return scanf("%lld",&a);}
int in(int &a){return scanf("%d",&a);}
int in(char *s){return scanf("%s",s);}
int in(char &s){return scanf("%c",&s);}
int in(db &a){return scanf("%Lf",&a);}
void out(int a){printf("%d ",a);}
void outln(int a){printf("%d\n",a);}
void out(ll a){printf("%lld ",a);}
void outln(ll a){printf("%lld\n",a);}
const db pi = acos((db)-1);
const ll inf =0x3f3f3f3f;
const db eps = 1e-8;
const int N = 5.1e5;
const ll mod = 998244353;
int sign(db a) { return a < -eps ? -1 : a > eps;}
int db_cmp(db a, db b){ return sign(a-b); }

int bg[N],g[N],en[N],h[N],a[N],b[N],len_h[N],len_a[N],cnt_h[N],cnt_a[N];//cnt[i]极大值树,len[i]
int lowbit(int x){
	return x&(-x);
}
void Update(int x,int y,int z,int n){
    if(len_a[x]==y){
        cnt_a[x]=(cnt_a[x]+z)%mod;
    }else{
        len_a[x]=y;;
        cnt_a[x]=z;
    }
    int lx;
    while(x<=n){
		len_h[x]=len_a[x];
        cnt_h[x]=cnt_a[x];
        lx = lowbit(x);
		for (int i=1; i<lx; i<<=1)
            if(len_h[x-i]>len_h[x]){
                len_h[x]=len_h[x-i];
                cnt_h[x]=cnt_h[x-i];
            }else if(len_h[x-i]==len_h[x])cnt_h[x]=(cnt_h[x]+cnt_h[x-i])%mod;
			// mx[x] = max(mx[x], len[x-i]); 
		x += lowbit(x);
    }
}
pii query(int l,int r){//返回 （以[l,r]结尾的最长LIS , 以[l,r]结尾的最长LIS的数量)  
    int ans = 0; 
    int tl=l,tr=r;
    while(r >= l){
      	ans = max(ans,len_a[r]);
      	r--;
		for(;r-lowbit(r)>= l ;r-=lowbit(r))ans = max(ans,len_h[r]);
    }
    l=tl;r=tr;
    int tot=0;
    while(r >= l){
      	// ans = max(ans,len_a[r]);
        if(len_a[r]==ans)tot=(tot+cnt_a[r])%mod;
      	r--;
		for(;r-lowbit(r)>= l ;r-=lowbit(r)){
            if(len_h[r]==ans)tot=(tot+cnt_h[r])%mod;
            // ans = max(ans,len_h[r]);
        }
    }
    return pii(ans,tot);
}
int qpow(ll a,ll b){
    ll ans=1;
    a%=mod;
    while(b){
        if(b%2)ans=ans*a%mod;
        a=a*a%mod;
        b/=2;
    }
    return ans;
}
int main() {
    int n;
    in(n);
    for0(i,n){
        in(a[i]);
        b[i]=a[i];
    }
    sort(b,b+n);
    for0(i,n)a[i]=lower_bound(b,b+n,a[i])-b+1;
    int L=0;
    pii tmp;
    for0(i,n){
        tmp=query(1,a[i]-1);
        en[i]=tmp.fi+1;
        L=max(L,en[i]);
        h[i]=tmp.se;
        if(en[i]==1)h[i]=1;
        Update(a[i],en[i],h[i],n);
    }
    mem0(cnt_a);mem0(cnt_h);mem0(len_a);mem0(len_h);
    int tot=0;
    for(int i=n-1;i>=0;i--){
        tmp=query(a[i]+1,n);
        bg[i]=tmp.fi+1;
        g[i]=tmp.se;
        if(bg[i]==1)g[i]=1;
        if(bg[i]==L)tot=(tot+g[i])%mod;
        Update(a[i],bg[i],g[i],n);
    }
    tot=qpow(tot,mod-2);
    for0(i,n){
        if(bg[i]+en[i]==L+1){
            out(1ll*h[i]*g[i]%mod*tot%mod);
        }else out(0);
    }
    return 0;
}