P3242 [HNOI2015] 接水果

题面

解法

这道题来回顾一个 trick，他就是可以将树上的操作问题通过 dfs 序转化为序列上的操作。

那么这道题我们对于苹果来找盘子是难做的，应为我对于全集找子集其实是困难的，但是我们对于子集找全集就变得简单了。而这道题我们看到看到第 \(k\) 大问题，我们想到的是二分答案，二分答案并判断小于等于这个答案的 \(mid\) 包含它的是否有 \(k\) 个这个使用扫描线加树状数组差分就可以了。

这道题有个教训，以后需要再编译的时候使用 -Werror 写程序一定要把所有的 warning 调到没有。

代码

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const ll N = 4e5 + 5;
struct node
{
    ll op, x, y, l, r, d, id;
} q[N], tl[N], tr[N];
ll ans[N], ntt;
ll c[N];
ll n, p, _;
inline ll lb(ll x)
{
    return x & (-x);
}
inline void add(ll x, ll d)
{
    while (x <= n)
    {
        c[x] += d;
        x += lb(x);
    }
}
inline ll query(ll x)
{
    ll ret = 0;
    while (x)
    {
        ret += c[x];
        x -= lb(x);
    }
    return ret;
}
bool cmp(node x, node y)
{
    if (x.x == y.x)
    {
        return x.op < y.op;
    }
    return x.x < y.x;
}
void solve(ll l, ll r, ll ql, ll qr)
{
    if (l > r)
    {
        return;
    }
    if (ql > qr)
    {
        return;
    }
    if (ql == qr)
    {
        for (int i = l; i <= r; i++)
        {
            if (q[i].op == 2)
            {
                ans[q[i].id] = ql;
            }
        }
        return;
    }
    ll mid = (ql + qr) >> 1;
    ll lt = 0, rt = 0;
    vector<node> qqq;
    for (int i = l; i <= r; i++)
    {
        if (q[i].op == 1)
        {
            if (q[i].d <= mid)
            {
                qqq.push_back({1, q[i].x, q[i].y, 0, 0, 1, 0});
                qqq.push_back({1, q[i].x, q[i].r + 1, 0, 0, -1, 0});
                qqq.push_back({1, q[i].l + 1, q[i].y, 0, 0, -1, 0});
                qqq.push_back({1, q[i].l + 1, q[i].r + 1, 0, 0, 1, 0});
                tl[++lt] = q[i];
            }
            else
            {
                tr[++rt] = q[i];
            }
        }
        else
        {
            qqq.push_back(q[i]);
        }
    }
    stable_sort(qqq.begin(), qqq.end(), cmp);
    for (auto &i : qqq)
    {
        if (i.op == 1)
        {
            add(i.y, i.d);
        }
        else
        {
            ll t = query(i.y);
            if (t < i.d)
            {
                i.d -= t;
                tr[++rt] = i;
            }
            else
            {
                tl[++lt] = i;
            }
        }
    }
    for (int i = 1; i <= lt; i++)
    {
        q[l + i - 1] = tl[i];
    }
    for (int i = 1; i <= rt; i++)
    {
        q[l + lt + i - 1] = tr[i];
    }
    if (lt)
    {
        solve(l, l + lt - 1, ql, mid);
    }
    if (rt)
    {
        solve(l + lt, r, mid + 1, qr);
    }
}
vector<ll> g[N];
ll st[N], ed[N], f[N][22], tot, dep[N];
void dfs(ll x, ll fa)
{
    f[x][0] = fa;
    for (int i = 1; i <= 20; i++)
    {
        f[x][i] = f[f[x][i - 1]][i - 1];
    }
    st[x] = ++tot;
    dep[x] = dep[fa] + 1;
    for (auto v : g[x])
    {
        if (v == fa)
        {
            continue;
        }
        dfs(v, x);
    }
    ed[x] = tot;
}
inline ll lca(ll x, ll y)
{
    if (dep[x] < dep[y])
    {
        swap(x, y);
    }
    ll cha = dep[x] - dep[y];
    for (int i = 0; i <= 20; i++)
    {
        if (cha >> i & 1)
        {
            x = f[x][i];
        }
    }
    if (x == y)
    {
        return x;
    }
    for (int i = 20; i >= 0; i--)
    {
        if (f[x][i] != f[y][i])
        {
            x = f[x][i], y = f[y][i];
        }
    }
    return f[x][0];
}
int main()
{
    // freopen("fruit1.in", "r", stdin);
    // freopen("fruit.out", "w", stdout);
    ios::sync_with_stdio(0), cin.tie(0);
    cin >> n >> p >> _;
    for (int i = 1; i < n; i++)
    {
        ll u, v;
        cin >> u >> v;
        g[u].push_back(v);
        g[v].push_back(u);
    }
    dfs(1, 0);
    for (int i = 1; i <= p; i++)
    {
        ll u, v, k;
        cin >> u >> v >> k;
        if (dep[u] > dep[v])
        {
            swap(u, v);
        }
        ll lc = lca(u, v);
        ll stu = st[u], edu = ed[u], stv = st[v], edv = ed[v];
        if (lc == u)
        {
            ll cha = dep[v] - dep[u] - 1;
            ll t = v;
            for (int j = 20; j >= 0; j--)
            {
                if (cha >> j & 1)
                {
                    t = f[t][j];
                }
            }
            if (st[t] >= 2)
            {
                q[++ntt] = node{1, 1, stv, st[t] - 1, edv, k, 0};
            }
            if (ed[t] + 1 <= n)
            {
                q[++ntt] = node{1, ed[t] + 1, stv, n, edv, k, 0};
            }
        }
        else
        {
            q[++ntt] = node{1, stu, stv, edu, edv, k, 0};
        }
        if (q[ntt].x > q[ntt].y)
        {
            swap(q[ntt].x, q[ntt].y);
            swap(q[ntt].l, q[ntt].r);
        }
    }
    for (int i = 1; i <= _; i++)
    {
        ll l, r, k;
        cin >> l >> r >> k;
        if (st[l] > st[r])
        {
            swap(l, r);
        }
        q[++ntt] = node{2, st[l], st[r], 0, 0, k, i};
    }
    solve(1, ntt, 0, 1e9);
    for (int i = 1; i <= _; i++)
    {
        cout << ans[i] << "\n";
    }
    return 0;
}