Submission #491137

#	Time	Username	Problem	Language	Result	Execution time	Memory
491137		ThaiBaHung	Beads and wires (APIO14_beads)	C++14	0 / 100	5 ms	5000 KiB

beads

#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 3;
int n;
int dp[N][2][2];
typedef pair < int, int > ii;
vector < ii > adj[N];

void dfs(int i, int j)
{
    dp[i][0][0] = 0;
    dp[i][0][1] = 0;
    //if (adj[i].size() == 2) dp[i][1][1] = 0;

    for (ii cur : adj[i])
    {
        int u = cur.first;
        int w = cur.second;
        if (u == j) continue;
        dfs(u, i);

        dp[i][1][1] = max(dp[i][1][1] + max({dp[u][0][0], dp[u][0][1], dp[u][1][0] + w}),
                          dp[i][1][0] + max({dp[u][1][1], dp[u][0][0], dp[u][0][1]}) + w);

        dp[i][1][0] = max(dp[i][1][0], dp[i][1][0] + max({dp[u][0][0], dp[u][0][1], dp[u][1][0] + w}));

        dp[i][1][0] = max(dp[i][1][0], dp[i][0][0] + max({dp[u][1][1], dp[u][0][0], dp[u][0][1]}) + w);

        dp[i][1][0] = max(dp[i][1][0], dp[i][0][1] + max({dp[u][1][1], dp[u][0][0], dp[u][0][1]}) + w);

        dp[i][0][0] += max({dp[u][0][0], dp[u][1][1], dp[u][0][1]});
        dp[i][0][1] += max({dp[u][1][1], dp[u][0][0], dp[u][0][1], dp[u][1][0] + w});

        //if (i == 1) cout << u << " " << dp[i][0][0] << " " << dp[i][0][1] << '\n';
    }

    //if (i == 10) cout << i << " " << dp[i][0][0] << " " << dp[i][0][1] << " " << dp[i][1][0] << " " << dp[i][1][1] << '\n';
}

int main()
{
    ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
    //freopen("t.inp","r",stdin); freopen("t.out","w",stdout);

    cin >> n;
    for (int i = 1; i < n; i++)
    {
        int u, v, w;
        cin >> u >> v >> w;
        adj[u].push_back({v, w});
        adj[v].push_back({u, w});
    }

    for (int i = 1; i <= n; i++)
        dp[i][1][1] = dp[i][1][0] = -1e9,
        dp[i][0][1] = dp[i][0][0] = -1e9;

    dfs(1, 0);

//    for (int i = 1; i <= n; i++)
//        cout << i << " " << dp[i][0][0] << " " << dp[i][0][1] << " " << dp[i][1][0] << " " << dp[i][1][1] << '\n';

    cout << max({dp[1][0][0], dp[1][0][1], dp[1][1][1]});
}

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