Submission #1057056

#	Time	Username	Problem	Language	Result	Execution time	Memory
1057056		fv3	Shortcut (IOI16_shortcut)	C++14	0 / 100	0 ms	348 KiB

shortcut

#include "shortcut.h"

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;

vector<vector<pair<int, ll>>> adj;
int N;

vector<int> visited;
vector<ll> dist;

ll find_diameter()
{
    visited = vector<int>(N);
    dist = vector<ll>(N, 1ll << 60);

    priority_queue<pair<ll, int>> q;
    q.push({0, 0});
    dist[0] = 0;

    int mx_index;
    while (1)
    {
        int s = q.top().second;
        q.pop();
        if (visited[s]) continue;
        visited[s] = 1;

        for (auto node : adj[s])
        {
            if (visited[node.first] || dist[s] + node.second > dist[node.first]) continue;
            dist[node.first] = dist[s] + node.second;
            q.push({-dist[node.first], node.first});
        }

        if (q.empty())
        {
            mx_index = s;
            break;
        }
    }

    visited = vector<int>(N);
    dist = vector<ll>(N, 1ll << 60);

    q.push({0, mx_index});
    dist[mx_index] = 0;

    int cnt = 0;
    while (1)
    {
        int s = q.top().second;
        q.pop(); 
        if (visited[s]) continue;
        visited[s] = 1; cnt++;

        for (auto node : adj[s])
        {
            if (visited[node.first] || dist[s] + node.second > dist[node.first]) continue;
            dist[node.first] = dist[s] + node.second;
            q.push({-dist[node.first], node.first});
        }

        if (cnt == N)
        {
            return dist[s];
        }
    }

    return 0;
}

ll find_shortcut(int n, std::vector<int> l, std::vector<int> d, int c)
{
    N = 2 * n;
    adj.resize(N);

    for (int i = 0; i < n - 1; i++)
    {
        adj[i].push_back({i+1, l[i]});
        adj[i+1].push_back({i, l[i]});
    }

    for (int i = 0; i < n; i++)
    {
        adj[i].push_back({i+n, d[i]});
        adj[i+n].push_back({i, d[i]});
    }

    ll res = 1ll << 60;
    for (int i = 0; i < n - 1; i++)
    {
        for (int j = i+1; j < n; j++)
        {
            adj[i].push_back({j, c});
            adj[j].push_back({i, c});

            res = min(res, find_diameter());

            adj[i].pop_back();
            adj[j].pop_back();
        }
    }

    return res;
}

• Subtask 1: 0 / 9
• Subtask 2: 0 / 14
• Subtask 3: 0 / 8
• Subtask 4: 0 / 7
• Subtask 5: 0 / 33
• Subtask 6: 0 / 22
• Subtask 7: 0 / 4
• Subtask 8: 0 / 3