제출 #380042

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
380042		Return_0	Dynamic Diameter (CEOI19_diameter)	C++17	100 / 100	630 ms	80524 KiB

diameter

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

#pragma GCC optimize ("Ofast")
// #pragma GCC target ("avx,avx2,fma")

using namespace std;
using namespace __gnu_pbds;

typedef long long ll;
typedef long double ld;
typedef pair <ll, ll> pll;

#ifdef LOCAL
#define dbg(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1> void __f(const char* name, Arg1&& arg1) { cout << name << " : " << arg1 << std::endl; }
template <typename Arg1, typename... Args> void __f (const char* names, Arg1&& arg1, Args&&... args) {
    const char* comma = strchr(names + 1, ',');cout.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...); }
#define dbg2(x, j, n) cout<< #x << " : "; output((j), (n), x, 1);
#else
#define dbg(...) 0
#define dbg2(x, j, n) 0
#endif
#define SZ(x) ((ll)((x).size()))
#define File(s, t) freopen(s ".txt", "r", stdin); freopen(t ".txt", "w", stdout);
#define input(j, n, a) for (int _i = (j); _i < (n)+(j); _i++) cin>> a[_i];
#define output(j, n, a, t) for (int _i = (j); _i < (n)+(j); _i++) cout<< a[_i] << (((t) && _i != (n)+(j)-1)? ' ' : '\n');
#define kill(x) return cout<< x << endl, 0
#define cl const ll
#define fr first
#define sc second
#define mid ((l + r) >> 1)
#define All(x) (x).begin(), (x).end()

const long long inf = 1e18;
cl mod = 1e9 + 7, MOD = 998244353;

template <class A, class B> ostream& operator << (ostream& out, const pair<A, B> &a) {
    return out << '(' << a.first << ", " << a.second << ')';    }
template <class A> ostream& operator << (ostream& out, const vector<A> &a) {
    if(!a.size())return cout<<"[]";cout<< '[' << a[0]; for (int i = 0; ++i < (int)(a.size());) cout<< ", " << a[i];cout<< ']';  }
template <class T, typename _t = less <T> > using Tree = tree <T, null_type, _t, rb_tree_tag, tree_order_statistics_node_update>;

cl N = 1e5 + 7;

ll tin [N], tout [N], e1 [N], e2 [N], W [N];
vector <ll> adj [N], walk;

void dfs (ll v = 1, ll par = 0) {
    walk.push_back(v);
    tin[v] = SZ(walk);
    for (auto u : adj[v]) if (u != par) {
        dfs(u, v);
        walk.push_back(v);
    }

    walk.push_back(v);
    tout[v] = SZ(walk);
}

struct node
{
    ll m, M, Mm, mM, ans;
    friend node operator + (const node &x, const node &y) {
        return {min(x.m, y.m), max(x.M, y.M), max({x.Mm, y.Mm, x.M - 2 * y.m}), max({x.mM, y.mM, y.M - 2 * x.m}), max({x.ans, y.ans, x.Mm + y.M, x.M + y.mM})};
    }
};

struct seg
{
    node n;
    ll lz, l, r;
    seg *lt, *rt;

    seg (ll L = 1, ll R = SZ(walk)) {
        n = {0, 0, 0, 0, 0};
        lz = 0, l = L, r = R;
    }

    void build () {
        if (l == r) return;

        lt = new seg (l, mid);
        rt = new seg (mid + 1, r);

        lt->build();
        rt->build();
    }

    inline void push (ll x) {   n.m += x; n.M += x; n.Mm -= x; n.mM -= x; lz += x;  }

    inline void shift () {  lt->push(lz);   rt->push(lz);   lz = 0; }

    void upd (ll ql, ll qr, ll x) {
        if (r < ql || qr < l)   return;

        if (ql <= l && r <= qr) {
            push(x);
            return;
        }

        shift();

        lt->upd(ql, qr, x);
        rt->upd(ql, qr, x);

        n = lt->n + rt->n;
    }
} mytree;

int main ()
{
    ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);

    ll n, q, mx_w, i, ans = 0, x, y;

    cin>> n >> q >> mx_w;
    for (i = 1; i < n; i++) {
        cin>> e1[i] >> e2[i] >> W[i];
        adj[e1[i]].push_back(e2[i]);
        adj[e2[i]].push_back(e1[i]);
    }

    dfs();

    mytree.r = SZ(walk);
    mytree.build();
    for (i = 1; i < n; i++) {
        if (tin[e1[i]] > tin[e2[i]])    swap(e1[i], e2[i]);
        mytree.upd(tin[e2[i]], tout[e2[i]], W[i]);
    }

    while (q--) {
        cin>> x >> y;
        x = (x + ans) % (n-1) + 1;
        y = (y + ans) % mx_w;

        mytree.upd(tin[e2[x]], tout[e2[x]], y - W[x]);
        W[x] = y;

        cout<< (ans = mytree.n.ans) << '\n';
    }

    cerr<< "\nTime elapsed: " << 1000 * clock() / CLOCKS_PER_SEC << "ms\n";

    return 0;
}
/*

*/

컴파일 시 표준 에러 (stderr) 메시지

diameter.cpp: In function 'std::ostream& operator<<(std::ostream&, const std::vector<_Tp>&)':
diameter.cpp:42:5: warning: this 'if' clause does not guard... [-Wmisleading-indentation]
   42 |     if(!a.size())return cout<<"[]";cout<< '[' << a[0]; for (int i = 0; ++i < (int)(a.size());) cout<< ", " << a[i];cout<< ']';  }
      |     ^~
diameter.cpp:42:36: note: ...this statement, but the latter is misleadingly indented as if it were guarded by the 'if'
   42 |     if(!a.size())return cout<<"[]";cout<< '[' << a[0]; for (int i = 0; ++i < (int)(a.size());) cout<< ", " << a[i];cout<< ']';  }
      |                                    ^~~~

• Subtask 1: 11 / 11
• Subtask 2: 13 / 13
• Subtask 3: 7 / 7
• Subtask 4: 18 / 18
• Subtask 5: 24 / 24
• Subtask 6: 27 / 27