这题用fhqtreap可以在线。
fhqtreap上维护以i结尾的最长上升子序列，数字按从小到大加入， 因为前面的数与新加入的数无关， 后面的数比新加入的数小， 所以新加入的数对原序列其他数的值没有影响。
然后就可以直接fhqtreap模拟啦
忘记所有答案取max WA了半天T T
#include<iostream> #include<cstring> #include<cstdlib> #include<cstdio> #define lt tree[x].ls #define rt tree[x].rs using namespace std; const int maxn=500010, seed=233333, inf=1e9+1; struct treap{int sum, mx, rnd, size, ls, rs;}tree[maxn]; int n, x, tott, root, ans; void read(int &k) { int f=1; k=0; char c=getchar(); while(c<'0' || c>'9') c=='-' && (f=-1), c=getchar(); while(c<='9' && c>='0') k=k*10+c-'0', c=getchar(); k*=f; } inline void build(int &x) { tree[x=++tott].rnd=rand()<<15|rand(); tree[x].sum=tree[x].mx=tree[x].size=1; } inline int max(int a, int b){return a>b?a:b;} inline void up(int x) { if(!x) return; tree[x].size=tree[lt].size+tree[rt].size+1; tree[x].mx=max(tree[x].sum, max(tree[lt].mx, tree[rt].mx)); } void split(int x, int &l, int &r, int k) { if(!k) l=0, r=x; else if(k==tree[x].size) l=x, r=0; else if(k<=tree[lt].size) r=x, split(lt, l, lt, k), up(x); else l=x, split(rt, rt, r, k-tree[lt].size-1), up(x); } void merge(int &x, int l, int r) { if(!l || !r) x=l+r; else if(tree[l].rnd<tree[r].rnd) x=l, merge(rt, rt, r), up(x); else x=r, merge(lt, l, lt), up(x); } inline int solve(int pos) { int x, y, tmp; build(tmp); split(root, x, y, pos); tree[tmp].sum=tree[tmp].mx=max(0, tree[x].mx)+1; merge(x, x, tmp); merge(root, x, y); return ans=max(ans, tree[tmp].sum); } int main() { srand(seed); read(n); tree[0].mx=-inf; tree[0].rnd=inf; for(int i=1;i<=n;i++) read(x), printf("%d\n", solve(x)); }