Submission #1002268

#	Time	Username	Problem	Language	Result	Execution time	Memory
1002268		GrindMachine	Beads and wires (APIO14_beads)	C++17	28 / 100	1027 ms	11732 KiB

beads

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
 
using namespace std;
using namespace __gnu_pbds;
 
template<typename T> using Tree = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
typedef long long int ll;
typedef long double ld;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
 
#define fastio ios_base::sync_with_stdio(false); cin.tie(NULL)
#define pb push_back
#define endl '\n'
#define sz(a) (int)a.size()
#define setbits(x) __builtin_popcountll(x)
#define ff first
#define ss second
#define conts continue
#define ceil2(x,y) ((x+y-1)/(y))
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
#define yes cout << "Yes" << endl
#define no cout << "No" << endl
 
#define rep(i,n) for(int i = 0; i < n; ++i)
#define rep1(i,n) for(int i = 1; i <= n; ++i)
#define rev(i,s,e) for(int i = s; i >= e; --i)
#define trav(i,a) for(auto &i : a)
 
template<typename T>
void amin(T &a, T b) {
    a = min(a,b);
}
 
template<typename T>
void amax(T &a, T b) {
    a = max(a,b);
}
 
#ifdef LOCAL
#include "debug.h"
#else
#define debug(x) 42
#endif
 
/*
 
read some solution ideas, found them non-intuitive 
trying to build upon my previous idea 

*/
 
const int MOD = 1e9 + 7;
const int N = 2e5 + 5;
const int inf1 = int(1e9) + 5;
const ll inf2 = ll(1e18) + 5;
 
vector<pii> adj[N];
vector<int> dp[N];
vector<int> best(N,-inf1);

void dfs(ll u, ll p){
    for(auto [v,w] : adj[u]){
        if(v == p) conts;
        dfs(v,u);
    }
 
    ll sum = 0;
 
    rep(i,sz(adj[u])){
        auto [v,w] = adj[u][i];
        if(v == p) conts;
        
        // ll curr_val = -inf2;
 
        // rep(j,sz(adj[v])){
        //     auto [v2,w2] = adj[v][j];
        //     if(v2 == u) conts;
 
        //     ll val = w+w2+dp[v2][0]+dp[v][j+1];
        //     amax(curr_val,val);            
        // }
        
        // amax(curr_val,dp[v][0]);

        ll curr_val = max(w+best[v],dp[v][0]);     
        dp[u][i+1] = -curr_val;
        sum += curr_val;
    }
 
    rep(i,sz(adj[u])+1){
        dp[u][i] += sum;
    }

    best[u] = -inf1;
    rep(j,sz(adj[u])){
        auto [v,w] = adj[u][j];
        if(v == p) conts;
        amax(best[u],w+dp[u][j+1]+dp[v][0]);
    }
}
 
void solve(int test_case)
{
    int n; cin >> n;
    rep1(i,n-1){
        int u,v,c; cin >> u >> v >> c;
        adj[u].pb({v,c}), adj[v].pb({u,c});
    }
    
    int ans = -inf1;    
    rep1(u,n){
        dp[u] = vector<int>(sz(adj[u])+1);
    }

    rep1(i,n){
        vector<pii> ord;
        queue<pii> q;
        q.push({i,-1});
        while(!q.empty()){
            auto [u,p] = q.front();
            q.pop();
            ord.pb({u,p});
            for(auto [v,w] : adj[u]){
                if(v == p) conts;
                q.push({v,u});
            }
        }

        reverse(all(ord));

        for(auto [u,p] : ord){
            int sum = 0;
            
            rep(i,sz(adj[u])){
                auto [v,w] = adj[u][i];
                if(v == p) conts;
                
                // ll curr_val = -inf2;
         
                // rep(j,sz(adj[v])){
                //     auto [v2,w2] = adj[v][j];
                //     if(v2 == u) conts;
         
                //     ll val = w+w2+dp[v2][0]+dp[v][j+1];
                //     amax(curr_val,val);            
                // }
                
                // amax(curr_val,dp[v][0]);

                int curr_val = max(w+best[v],dp[v][0]);     
                dp[u][i+1] = -curr_val;
                sum += curr_val;
            }
            
            dp[u][0] = 0;
            rep(i,sz(adj[u])+1){
                dp[u][i] += sum;
            }

            best[u] = -inf1;
            rep(j,sz(adj[u])){
                auto [v,w] = adj[u][j];
                if(v == p) conts;
                amax(best[u],w+dp[u][j+1]+dp[v][0]);
            }
        }

        amax(ans,dp[i][0]);

        // dfs(i,-1);
        // amax(ans,dp[i][0]);
    }
 
    cout << ans << endl;
}
 
int main()
{
    fastio;
 
    int t = 1;
    // cin >> t;
 
    rep1(i, t) {
        solve(i);
    }
 
    return 0;
}

• Subtask 1: 13 / 13
• Subtask 2: 15 / 15
• Subtask 3: 0 / 29
• Subtask 4: 0 / 43