cf1132G. Greedy Subsequences(线段树)

题意

题目链接

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;
}