제출 #563579

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
563579		piOOE	구슬과 끈 (APIO14_beads)	C++17	100 / 100	154 ms	23996 KiB

beads

//#define _GLIBCXX_DEBUG

//#pragma GCC optimize("Ofast")
//#pragma GCC optimize("unroll-loops")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")

#include <bits/stdc++.h>

using namespace std;

//#include <ext/pb_ds/assoc_container.hpp>
//
//using namespace __gnu_pbds;
//
//template<typename T>
//using ordered_set = tree<T, null_type, less < T>, rb_tree_tag, tree_order_statistics_node_update>;

//template<typename T>
//using normal_queue = priority_queue<T, vector<T>, greater<>>;

mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());

#define trace(x) cout << #x << ": " << (x) << endl;
#define all(x) begin(x), end(x)
#define rall(x) rbegin(x), rend(x)
#define uniq(x) x.resize(unique(all(x)) - begin(x))
#define sz(s) ((int) size(s))
#define pii pair<int, int>
#define mp(x, y) make_pair(x, y)
#define int128 __int128
#define pb push_back
#define popb pop_back
#define eb emplace_back
#define fi first
#define se second
#define itn int

typedef long long ll;
typedef pair<ll, ll> pll;
typedef long double ld;
typedef double db;
typedef unsigned int uint;


template<typename T>
bool ckmn(T &x, T y) {
    if (x > y) {
        x = y;
        return true;
    }
    return false;
}

template<typename T>
bool ckmx(T &x, T y) {
    if (x < y) {
        x = y;
        return true;
    }
    return false;
}

int bit(int x, int b) {
    return (x >> b) & 1;
}

int rand(int l, int r) { return (int) ((ll) rnd() % (r - l + 1)) + l; }


const ll infL = 3e18;
const int infI = 1'000'000'000 + 7;
const int infMI = 0x3f3f3f3f; //a little bigger than 1e9
const ll infML = 0x3f3f3f3f3f3f3f3fLL; //4.5e18
const int N = 200002;
const int mod = 998244353;
const ld eps = 1e-9;

vector<pair<int, int>> g[N];

int dp[3][N];
//dp[0][v] - just simple, if we dont merge sons in our subtree and also not use our dad
//dp[1][v] - also if we dont merge sons in our subtree, but use our dad
//dp[2][v] - answer

int p[N], wp[N];
//wp - weight of edge to our parent

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int n;
    cin >> n;
    for (int i = 1; i < n; ++i) {
        int a, b, w;
        cin >> a >> b >> w;
        --a, --b;
        g[a].emplace_back(b, w);
        g[b].emplace_back(a, w);
    }
    p[0] = -1;
    function<void(int)> dfs = [&dfs](int v) {
        int mx1 = -infI, mx2 = -infI;
        int sum = 0;
        for (auto [to, w]: g[v]) {
            if (to != p[v]) {
                p[to] = v;
                wp[to] = w;
                dfs(to);
                int mx_now = max(dp[0][to], dp[1][to]);
                sum += mx_now;
                int val = dp[0][to] + w - mx_now;
                if (val > mx1) {
                    mx2 = mx1;
                    mx1 = val;
                } else if (mx2 < val){
                    mx2 = val;
                }
            }
        }
        dp[0][v] = dp[2][v] = sum;
        for (auto [to, w]: g[v]) {
            if (to != p[v]) {
                int mx_now = max(dp[0][to], dp[1][to]);
                int val = dp[0][v] - mx_now;
                if (p[v] != -1) {
                    ckmx(dp[1][v], dp[0][to] + w + val + wp[v]);
                }
                int mx = w + dp[0][to] - mx_now;
                if (mx == mx1) {
                    mx = mx2;
                } else {
                    mx = mx1;
                }
                ckmx(dp[2][v], dp[2][to] + val + mx + w);
                ckmx(dp[2][v], dp[2][to] + val);
                for (auto [u, k] : g[to]) {
                    if (u != p[to]) {
                        int valu = dp[0][to] - max(dp[0][u], dp[1][u]);
                        ckmx(dp[2][v], dp[2][u] + k + w + val + valu);
                    }
                }
            }
        }
    };
    dfs(0);
    cout << dp[2][0];
    return 0;
}

• Subtask 1: 13 / 13
• Subtask 2: 15 / 15
• Subtask 3: 29 / 29
• Subtask 4: 43 / 43