Submission #1001301

#	Time	Username	Problem	Language	Result	Execution time	Memory
1001301		LonlyR	Tourism (JOI23_tourism)	C++17	28 / 100	5050 ms	53676 KiB

tourism

#include<bits/stdc++.h>
#define int signed
#define all(x) x.begin(), x.end()
#define ii pair<int,int>
#define ff first
#define ss second
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
 
using namespace std;
const int maxn = 1e5 + 5, bl = sqrt(maxn);
int n, m, q, cnt, cur;
int in[maxn], out[maxn], h[maxn], par[18][maxn], id[maxn], c[maxn], ans[maxn], in_set[maxn];
vector<int> adj[maxn], euler = {0};
int lgs[3*maxn];
 
int rev_in[maxn];
 
set<int> s;
 
struct Q{
    int l, r, id;
    bool operator < (Q b) const
    {
        if (l / bl != b.l / bl) return l < b.l;
        return (l / bl & 1) ? (r < b.r) : (r > b.r);
    }
} qr[maxn];
 
void dfs(int x = 1, int p = 1)
{
    h[x] = h[p] + 1;
    in[x] = ++cnt;
    id[x] = euler.size();
    euler.push_back(x);
    for (int i : adj[x]) if (i != p)
        dfs(i, x),
        euler.push_back(x);
    out[x] = cnt;
}
 
//inline bool sub(int u, int v) /// u in subtree of v
//{
//    return in[v] <= in[u] && out[u] <= out[v];
//}
//
//int lca(int u, int v)
//{
//    if (h[u] > h[v]) swap(u, v);
//    if (sub(v, u)) return u;
//    for (int i = 17; i >= 0; i--)
//        if (sub(u, par[i][v]) == 0)
//            v = par[i][v];
//    return par[0][v];
//}
 
vector<vector<ii>> st;
 
inline void build()
{
    int sz = euler.size();
    int LG = __lg(sz);
    st = vector<vector<ii>>(LG + 1, vector<ii>(sz, {sz, 0}));
    for (int i = 1; i < sz; i++) st[0][i] = {h[euler[i]], euler[i]};
    for (int j = 1; j <= LG; j++)
        for (int i = 1; i + (1 << j) - 1 < sz; i++)
            st[j][i] = min(st[j - 1][i], st[j - 1][i + (1 << (j - 1))]);
}
 
inline int lca(int l, int r)
{
    l = id[l];
    r = id[r];
    if (l > r) swap(l, r);
//    assert(r-l+1<=n);
    int k = lgs[(r - l + 1)];
    return min(st[k][l], st[k][r - (1 << k) + 1]).ss;
}
 
inline void add(int x)
{
    x = c[x];
    in_set[x]++;
    if (in_set[x] == 1)
    {
        cur += h[x];
        if (s.size() == 0) return s.insert(in[x]), void();
        auto it = s.lower_bound(in[x]);
        if (it == s.begin())
            cur -= h[lca(x, rev_in[*it])];
        else if (it == s.end())
            cur -= h[lca(x, rev_in[*prev(it)])];
        else
        {
            auto u = rev_in[*it], v = rev_in[*prev(it)];
            cur += h[lca(u, v)];
            cur -= h[lca(u, x)];
            cur -= h[lca(v, x)];
        }
    }
    s.insert(in[x]);
}
 
inline void del(int x)
{
    x = c[x];
    in_set[x]--;
    if (in_set[x] == 0)
    {
        cur -= h[x];
        if (s.size() == 1) return s.clear(), void();
        auto it = s.lower_bound(in[x]);
        if (it == s.begin())
            cur += h[lca(x, rev_in[*next(it)])];
        else if (next(it) == s.end())
            cur += h[lca(x, rev_in[*prev(it)])];
        else
        {
            auto u = rev_in[*prev(it)], v = rev_in[*next(it)];
            cur -= h[lca(u, v)];
            cur += h[lca(u, x)];
            cur += h[lca(v, x)];
        }
        s.erase(it);
    }
 
}
 
signed main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL); //cout.tie(0);
//    freopen("test.inp", "r", stdin);
//    freopen("test.out", "w", stdout);
    cin >> n >> m >> q;
    for (int i = 1, u, v; i < n; i++)
        cin >> u >> v,
        adj[u].push_back(v),
        adj[v].push_back(u);
    for (int i = 1; i <= m; i++)
        cin >> c[i];
    dfs();
    for (int i = 1; i <= 3*n; i++) lgs[i] = __lg(i);
    build();
    for (int i = 1; i <= n; i++) rev_in[in[i]] = i;
    for (int i = 1; i <= q; i++)
        cin >> qr[i].l >> qr[i].r, qr[i].id = i;
    sort(qr + 1, qr + q + 1);
    for (int i = 1, l = 1, r = 0; i <= q; i++)
    {
        auto [lx, rx, id] = qr[i];
        while (r < rx) add(++r);
        while (l > lx) add(--l);
        while (r > rx) del(r--);
        while (l < lx) del(l++);
        ans[id] = cur - h[lca(rev_in[*s.begin()], rev_in[*s.rbegin()])] + 1;
    }
    for (int i = 1; i <= q; i++) cout << ans[i] << "\n";
}

• Subtask 1: 5 / 5
• Subtask 2: 5 / 5
• Subtask 3: 0 / 7
• Subtask 4: 18 / 18
• Subtask 5: 0 / 24
• Subtask 6: 0 / 41