Greedy Subsequences CodeForces - 1132G

我们从右往左滑动区间, 假设dp[i]表示i为左端点时的最大长度, 通过观察可以发现, 每添加一个点, 该点$dp$值=它右侧第一个比它大位置处$dp$值+1, 但是每删除一个点会将所有以它为根的$dp$值全-1, 所以可以根据转移建一棵树, 需要有单点查询单点更新以及树链加, 可以用线段树维护dfs序$O(logn)$实现, 或者直接树剖$O(nlog^2n)$

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <math.h>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <string.h>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,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'
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, 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;}
//head
#ifdef ONLINE_JUDGE
const int N = 1e6+10;
#else
const int N = 111;
#endif

int n, k;
vector<int> g[N], q;
int a[N], L[N], R[N], dp[N], fa[N], ans[N];
int val[N<<2], tag[N<<2];
int no[N];

void dfs(int x) {
	L[x]=++*L,no[*L]=x;
	for (int y:g[x]) dfs(y);
	R[x]=*L;
}
void pd(int o) {
	if (tag[o]) {
		tag[lc]+=tag[o];
		tag[rc]+=tag[o];
		val[lc]+=tag[o];
		val[rc]+=tag[o];
		tag[o]=0;
	}
}
void upd1(int o, int l, int r, int ql, int qr, int v) {
	if (ql<=l&&r<=qr) return val[o]+=v,tag[o]+=v,void();
	pd(o);
	if (mid>=ql) upd1(ls,ql,qr,v);
	if (mid<qr) upd1(rs,ql,qr,v);
	val[o]=max(val[lc],val[rc]);
}
void upd2(int o, int l, int r, int x, int v) {
	if (l==r) return val[o]=v,void();
	pd(o);
	if (mid>=x) upd2(ls,x,v);
	else upd2(rs,x,v);
	val[o]=max(val[lc],val[rc]);
}
int qry(int o, int l, int r, int x) {
	if (l==r) return val[o];
	pd(o);
	if (mid>=x) return qry(ls,x);
	return qry(rs,x);
}
void dfs(int o, int l, int r) {
	if (l==r) printf("no=%d,a=%d,val=%d\n",no[l],a[no[l]],val[o]);
	else pd(o),dfs(ls),dfs(rs);
}

int main() {
	scanf("%d%d", &n, &k);
	REP(i,1,n) scanf("%d", a+i);
	a[++n] = INF;
	q.pb(n);
	PER(i,1,n-1) {
		while (a[i]>=a[q.back()]) q.pop_back();
		int j = q.back();
		g[j].pb(i), fa[i] = j, q.pb(i);
	}
	dfs(n);
	PER(i,1,n-1) {
		if (i+k<n) upd1(1,1,n,L[i+k],R[i+k],-1);
		if (fa[i]-i>=k) dp[i] = 1;
		else dp[i] = qry(1,1,n,L[fa[i]])+1;
		upd2(1,1,n,L[i],dp[i]);
		if (i+k<=n) ans[i]=val[1];
	}
	REP(i,1,n-k) printf("%d ", ans[i]);hr;
}