NOIP 模拟 $27\; \rm 牛半仙的妹子序列$

题解 \(by\;zj\varphi\)

明显一道极长上升子序列的题。

直接线段树维护单调栈，最后单调栈求出可以贡献的序列，答案相加就行。

Code

#include<bits/stdc++.h>
#define ri register signed
#define p(i) ++i
using namespace std;
namespace IO{
    char buf[1<<21],*p1=buf,*p2=buf,OPUT[100];
    #define gc() p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?(-1):*p1++;
    template<typename T>inline void read(T &x) {
        ri f=1;x=0;register char ch=gc();
        while(!isdigit(ch)) {if (ch=='-') f=0;ch=gc();}
        while(isdigit(ch)) {x=(x<<1)+(x<<3)+(ch^48);ch=gc();}
        x=f?x:-x;
    }
    template<typename T>inline void print(T x,char t) {
        if (x<0) putchar('-'),x=-x;
        if (!x) return putchar('0'),(void)putchar(t);
        ri cnt(0);
        while(x) OPUT[p(cnt)]=x%10,x/=10;
        for (ri i(cnt);i;--i) putchar(OPUT[i]^48);
        return (void)putchar(t);
    }
}
using IO::read;using IO::print;
namespace nanfeng{
    #define FI FILE *IN
    #define FO FILE *OUT
    template<typename T>inline T cmax(T x,T y) {return x>y?x:y;}
    template<typename T>inline T cmin(T x,T y) {return x>y?y:x;}
    static const int N=2e5+7,MOD=998244353;
    int pos[N],dp[N],v[N],st[N],wk[N],cnt,n,rmx,tmx,ans;
    inline int MD(int x) {return x>=MOD?x-MOD:x;}
    struct Seg{
        #define ls(x) (x<<1)
        #define rs(x) (x<<1|1)
        struct segmenttree{int sum,mx,tg;}T[N<<2];
        void build(int x,int l,int r) {
            if (l==r) return (void)(T[x].tg=1);
            int mid(l+r>>1);
            build(ls(x),l,mid);
            build(rs(x),mid+1,r);
        }
        int calc(int x,int mx) {
            if (T[x].mx<=mx) return 0;
            if (T[x].tg) return dp[T[x].mx];
            if (T[rs(x)].mx<mx) return calc(ls(x),mx);
            return MD(T[x].sum+calc(rs(x),mx));
        }
        inline void up(int x) {
            T[x].mx=cmax(T[ls(x)].mx,T[rs(x)].mx);
            T[x].sum=calc(ls(x),T[rs(x)].mx);
        }
        void update(int x,int k,int p,int l,int r) {
            if (l==r) return (void)(T[x].mx=k);
            int mid(l+r>>1);
            if (p<=mid) update(ls(x),k,p,l,mid);
            else update(rs(x),k,p,mid+1,r);
            up(x);
        }
        int query(int x,int l,int r,int lt,int rt) {
            if (l<=lt&&rt<=r) return tmx=rmx,rmx=cmax(rmx,T[x].mx),calc(x,tmx);
            int mid(lt+rt>>1),res(0);
            if (r>mid) res+=query(rs(x),l,r,mid+1,rt);
            if (l<=mid) res+=query(ls(x),l,r,lt,mid);
            return MD(res);
        }
    }T;
    inline int main() {
        //FI=freopen("nanfeng.in","r",stdin);
        //FO=freopen("nanfeng.out","w",stdout);
        read(n);
        for (ri i(1);i<=n;p(i)) read(v[i]),pos[v[i]]=i;
        ri tp=0;
        for (ri i(1);i<=n;p(i)) {
            while(tp&&v[st[tp]]<v[i]) --tp;
            st[p(tp)]=i;
        }
        while(tp) wk[p(cnt)]=st[tp--];
        T.build(1,1,n);
        for (ri i(1);i<=n;p(i)) {
            ri cur(pos[i]);
            rmx=0;
            dp[i]=T.query(1,1,cur,1,n);
            if (!dp[i]) dp[i]=1;
            T.update(1,i,cur,1,n);
        }
        for (ri i(1);i<=cnt;p(i)) ans=MD(ans+dp[v[wk[i]]]);
        print(ans,'\n');
        return 0;
    }
}
int main() {return nanfeng::main();}