题意
Sol
昨天没想到真是有点可惜了。
我们考虑每个点作为最大值的贡献，首先预处理出每个位置\(i\)左边第一个比他大的数\(l\)，显然\([l + 1, i]\)内的数的后继要么是\(i\)，要么在这一段区间中。那么可以对这段区间\(+1\)，然后每次查询\([i - k + 1, i]\)的最大值即可
#include<bits/stdc++.h>
#define Pair pair<int, int>
#define MP(x, y) make_pair(x, y)
#define fi first
#define se second
//#define int long long
#define LL long long
#define Fin(x) {freopen(#x".in","r",stdin);}
#define Fout(x) {freopen(#x".out","w",stdout);}
using namespace std;
const int MAXN = 4e6 + 10, mod = 1e9 + 7, INF = 1e9 + 10;
const double eps = 1e-9;
template <typename A, typename B> inline bool chmin(A &a, B b){if(a > b) {a = b; return 1;} return 0;}
template <typename A, typename B> inline bool chmax(A &a, B b){if(a < b) {a = b; return 1;} return 0;}
template <typename A, typename B> inline LL add(A x, B y) {if(x + y < 0) return x + y + mod; return x + y >= mod ? x + y - mod : x + y;}
template <typename A, typename B> inline void add2(A &x, B y) {if(x + y < 0) x = x + y + mod; else x = (x + y >= mod ? x + y - mod : x + y);}
template <typename A, typename B> inline LL mul(A x, B y) {return 1ll * x * y % mod;}
template <typename A, typename B> inline void mul2(A &x, B y) {x = (1ll * x * y % mod + mod) % mod;}
template <typename A> inline void debug(A a){cout << a << '\n';}
template <typename A> inline LL sqr(A x){return 1ll * x * x;}
inline int read() {
char c = getchar(); int x = 0, f = 1;
while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
return x * f;
}
int N, K, a[MAXN], st[MAXN], top, pre[MAXN];
int root, ls[MAXN], rs[MAXN], tag[MAXN], mx[MAXN], ll[MAXN], rr[MAXN], tot, times;
void ps(int &k, int v, int l, int r) {
if(!k) k = ++tot, ll[k] = l, rr[k] = r;
mx[k] += v;
tag[k] += v;
}
void pushdown(int k, int l, int r) {
if(!tag[k]) return ;
int mid = l + r >> 1;
ps(ls[k], tag[k], l, mid);
ps(rs[k], tag[k], mid + 1, r);
tag[k] = 0;
}
void update(int k) {
mx[k] = max(mx[ls[k]], mx[rs[k]]);
}
void IntAdd(int &k, int l, int r, int ql, int qr, int v) {
if(!k) k = ++tot, ll[k] = l, rr[k] = r;
if(ql <= l && r <= qr) {ps(k, v, l, r); return ;}
pushdown(k, l, r);
int mid = l + r >> 1;
if(ql <= mid) IntAdd(ls[k], l, mid, ql, qr, v);
if(qr > mid) IntAdd(rs[k], mid + 1, r, ql, qr, v);
update(k);
}
int Query(int k, int l, int r, int ql, int qr) {
if(!k) return 0;
if(ql <= l && r <= qr) return mx[k];
pushdown(k, l, r); int mid = l + r >> 1;
if(ql > mid) return Query(rs[k], mid + 1, r, ql, qr);
else if(qr <= mid) return Query(ls[k], l, mid, ql, qr);
else return max(Query(ls[k], l, mid, ql, qr), Query(rs[k], mid + 1, r, ql, qr));
}
signed main() {
N = read(); K = read();
for(int i = 1; i <= N; i++) a[i] = read();
for(int i = N; i; i--) {
while(top && a[st[top]] <= a[i]) pre[st[top--]] = i;
st[++top] = i;
}
while(top) pre[st[top--]] = 0;
for(int i = 1; i <= N; i++) {
IntAdd(root, 1, N, pre[i] + 1, i, 1);
if(i >= K)
cout << Query(root, 1, N, i - K + 1, i) << " ";
}
return 0;
}