Submission #344765

#	Time	Username	Problem	Language	Result	Execution time	Memory
344765		Sara	Beads and wires (APIO14_beads)	C++14	0 / 100	4 ms	5100 KiB

beads

#include <bits/stdc++.h>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
using namespace std;
using namespace __gnu_pbds;
template <class T> using Tree = tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update>;

#define ll long long int
#define F first
#define S second
#define pb push_back
#define v00 dp[0][0][v]
#define v01 dp[0][1][v]
#define v10 dp[1][0][v]
#define v11 dp[1][1][v]
#define u00 dp[0][0][u]
#define u01 dp[0][1][u]
#define u10 dp[1][0][u]
#define u11 dp[1][1][u]

const ll M = 5;
const ll N = 2e5 + 5;
const ll LOG = 20;
const int MOD = 1e9 + 7;
const ll INF = 1e9 + 5;

ll n, dp[M][M][N];
vector < pair <int, ll> > adj[N]; 

// dp[0][0][v] : (in sub[v] down : 0) && (v up_dn : 0) -> ok : v is emp
// dp[0][1][v] : (in sub[v] down : 0) && (v up_dn : 1) -> ok : v is up_dn
// dp[1][0][v] : (in sub[v] down : 1) && (v up_dn : 0) -> ok : v is dn || v is emp
// dp[1][1][v] : (in sub[v] down : 1) && (v up_dn : 1) -> ok : v is up_dn

void dfs(int v = 1, int p = 0, ll up = -INF){
    v00 = v01 = v10 = v11 = 0;
    ll hlp[] = {0, -INF, -INF, -INF};
    for (auto i : adj[v]){
        int u = i.F;
        ll dn = i.S;
        if (u == p) continue;
        dfs(u, v, i.S); 
        v11 = max({v11 + max(u00, u01), v01 + max(u10, u11), v10 + up + dn + u00, v00 + up + dn + u10});
        v10 = max({v10 + max(u00, u01), v00 + max(u10, u11)});
        v01 = max(v01 + max(u00, u01), v00 + up + dn + u00);
        v00 = v00 + max(u00, u01);
        hlp[3] = max({hlp[3] + max(u00, u01), hlp[2] + dn + u00, hlp[1] + dn + max(u00, u10)});
        hlp[2] = max(hlp[2] + max(u00, u01), hlp[0] + dn + max(u00, u10));
        hlp[1] = max(hlp[1] + max(u00, u01), hlp[0] + dn + u00);
        hlp[0] = hlp[0] + max(u00, u01);
    }
    v10 = max(v10, hlp[3]);
    return;
}

int main() {
    ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    cin >> n;
    for (int i = 0; i < n - 1; i ++){
        int u, v, w;
        cin >> u >> v >> w;
        adj[u].pb({v, w});
        adj[v].pb({u, w});
    }
    dfs();
    cout << max(dp[0][0][1], dp[1][0][1]) << '\n';
    return 0;
}

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