Submission #1002377

#	Time	Username	Problem	Language	Result	Execution time	Memory
1002377		GrindMachine	Beads and wires (APIO14_beads)	C++17	100 / 100	135 ms	35320 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 
built upon my initial idea for ac

*/
 
const int MOD = 1e9 + 7;
const int N = 2e5 + 5;
const int inf1 = int(1e9) + 5;
const ll inf2 = ll(1e18) + 5;

struct S : vector<int>{
    int shift = 0;

    int operator[](int i){
        return at(i)+shift;
    }

    void point_set(int i, int v){
        at(i) = v-shift;
    }

    void point_add(int i, int v){
        at(i) += v;
    }

    void do_shift(int x){
        shift += x;
    }
};

template<typename T>
struct two_max{
    array<T,2> ar;
    two_max(){
        ar.fill(T{-inf1,-inf1});
    }

    void insert(T px){
        if(px > ar[0]){
            ar[1] = ar[0];
            ar[0] = px;
        }
        else{
            amax(ar[1],px);
        }
    }
};

vector<pii> adj[N];
S dp[N];
vector<int> best(N,-inf1);

void dfs1(ll u, ll p){
    for(auto [v,w] : adj[u]){
        if(v == p) conts;
        dfs1(v,u);
    }
    
    dp[u].point_set(0,0);    
    int sum = 0;

    rep(i,sz(adj[u])){
        auto [v,w] = adj[u][i];
        if(v == p) conts;
        int curr_val = max(w+best[v],dp[v][0]);
        sum += curr_val;
        dp[u].point_set(i+1,-curr_val);
    }
    
    rep(i,sz(adj[u])+1){
        dp[u].point_add(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]);
    }
}

int ans = 0;

void dfs2(int u, int p){
    amax(ans,dp[u][0]);
    two_max<pii> vals;

    rep(j,sz(adj[u])){
        auto [v2,w2] = adj[u][j];
        int val = w2+dp[u][j+1]+dp[v2][0];
        vals.insert({val,v2});
    }

    rep(i,sz(adj[u])){
        auto [v,w] = adj[u][i];
        if(v == p) conts;

        int id = -1;
        rep(j,sz(adj[v])){
            auto [v2,w2] = adj[v][j];
            if(v2 == u){
                id = j+1;
                break;
            }
        }

        int contrib = max(w+best[v],dp[v][0]);
        dp[u].point_add(i+1,contrib);
        dp[u].do_shift(-contrib);

        int add = 0;
        rep(j,2){
            auto [val,v2] = vals.ar[j];
            if(v2 == v or v2 < 0) conts;
            amax(add,w+val-contrib);
        }

        // rep(j,sz(adj[u])){
        //     if(j == i) conts;
        //     auto [v2,w2] = adj[u][j];
        //     int val = w+w2+dp[u][j+1]+dp[v2][0];
        //     amax(add,val);
        // }
        
        amax(add,dp[u][0]);

        dp[v].point_add(id,-add);
        rep(j,sz(adj[v])+1){
            dp[v].point_add(j,add);
        }

        dfs2(v,u);

        dp[u].do_shift(contrib);
        dp[u].point_add(i+1,-contrib);

        rep(j,sz(adj[v])+1){
            dp[v].point_add(j,-add);
        }
        dp[v].point_add(id,add);
    }
}

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});
    }
    
    rep1(u,n){
        dp[u].resize(sz(adj[u])+1);
    }

    dfs1(1,-1);
    dfs2(1,-1);

    cout << ans << endl;

    // int oans = 0;
    // rep1(i,n){
    //     dfs1(i,-1);
    //     amax(oans,dp[i][0]);
    //     debug(dp[i][0]);
    // }

    // assert(ans == oans);
}
 
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: 29 / 29
• Subtask 4: 43 / 43