J. Joy of Handcraft

题意：

给定 n 个灯泡的时间周期以及对应的亮度值，求 1 ~ m 的时刻，每一时刻的灯泡最大亮度

分析：

按时间轴建树，维护时间区间的亮度最大值
按亮度值递减排序，遍历灯泡时只 modify 为相同周期中亮度值最大的一个灯泡作为区间亮度最大值
区间修改，单点查询

实现：

#include <bits/stdc++.h>
using namespace std;
#define mst(x, y) memset(x, y, sizeof x)
#define endl '\n'
#define INF LONG_LONG_MAX
#define int long long
#define Lson u << 1, l, mid
#define Rson u << 1 | 1, mid + 1, r
#define FAST ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
const int N = 200010, MOD = 1e9 + 7;
const double EPS = 1e-6;
typedef pair<int, int> PII;
int T;
int n, m;
int Case = 1;
bool st[N];
struct Q
{
    int t, x;
} q[N];
struct Node
{
    int l, r;
    int val;
    int lz;
} tr[N << 2];
void pushup(Node u, Node l, Node r)
{
    u.val = max(l.val, r.val);
}
void pushup(int u)
{
    pushup(tr[u], tr[u << 1], tr[u << 1 | 1]);
}
void pushdown(int u)
{
    tr[u << 1].lz = max(tr[u].lz, tr[u << 1].lz);
    tr[u << 1 | 1].lz = max(tr[u].lz, tr[u << 1 | 1].lz);
    tr[u << 1].val = max(tr[u << 1].val, tr[u].lz);
    tr[u << 1 | 1].val = max(tr[u << 1 | 1].val, tr[u].lz);
    tr[u].lz = 0;
}
void build(int u, int l, int r)
{
    if (l == r)
        tr[u] = {l, l, 0, 0};
    else
    {
        tr[u] = {l, r};
        int mid = l + r >> 1;
        build(Lson), build(Rson);
        pushup(u);
    }
}
void modify(int u, int l, int r, int k)
{
    if (tr[u].l >= l && tr[u].r <= r)
    {
        tr[u].lz = max(tr[u].lz, k);
        tr[u].val = max(tr[u].val, tr[u].lz);
    }
    else
    {
        pushdown(u);
        int mid = tr[u].l + tr[u].r >> 1;
        if (l <= mid)
            modify(u << 1, l, r, k);
        if (r > mid)
            modify(u << 1 | 1, l, r, k);
        pushup(u);
    }
}
int query(int u, int l, int r)
{
    if (tr[u].l >= l && tr[u].r <= r)
        return tr[u].val;
    else
    {
        pushdown(u);
        int mid = tr[u].l + tr[u].r >> 1;
        if (l <= mid)
            return query(u << 1, l, r);
        else
            return query(u << 1 | 1, l, r);
    }
}
void solve()
{
    mst(st, false);
    cin >> n >> m;
    for (int i = 1; i <= n; i++)
        cin >> q[i].t >> q[i].x;
    sort(q + 1, q + 1 + n, [&](Q a, Q b)
         { return a.x > b.x; });

    build(1, 1, m);

    for (int i = 1; i <= n; i++)
    {
        if (st[q[i].t])
            continue;
        st[q[i].t] = true;

        int l = 1, r = q[i].t;
        while (l <= m)
        {
            // int tmp = query(1, l, min(r, m));
            // if (tmp < q[i].x)
                modify(1, l, min(r, m), q[i].x);
            l += 2 * q[i].t, r += 2 * q[i].t;
        }
    }

    for (int i = 1; i <= m; i++)
    {
        cout << query(1, i, i) << " \n"[i == m];
    }
}
signed main()
{
    FAST;
    T = 1;
    cin >> T;
    while (T--)
    {
        cout << "Case #" << Case++ << ": ";
        solve();
    }
    return 0;
}