Submission #1271639

#	Time	Username	Problem	Language	Result	Execution time	Memory
1271639		cokalhado	Olympic Bus (JOI20_ho_t4)	C++20	0 / 100	1 ms	320 KiB

ho_t4

// https://oj.uz/problem/view/JOI20_ho_t4
#include<bits/stdc++.h>
using namespace std;

#define int long long

const int maxn = 210;
const int inf = 1e18;
const int comp = 1e17;
int N, M;
vector<pair<int, int>> g[maxn];
tuple<int, int, int, int> ar[maxn];
int dist[maxn];
bool proc[maxn];
int ans = inf;

void dijkstra(int A)
{
    for(int i = 1; i <= N; i++)
    {
        dist[i] = 0;
        proc[i] = 0;
    }
    priority_queue<pair<int, int>, vector<pair<int, int>>, greater<pair<int, int>>> pq;
    pq.push({1, A});
    dist[A] = 1;
    while(!pq.empty())
    {
        int x = pq.top().second;
        pq.pop();
        if(proc[x]) continue;
        proc[x] = 1;
        for(auto[v, w] : g[x])
        {
            if(dist[v] == 0 || dist[x]+w < dist[v])
            {
                dist[v] = dist[x]+w;
                pq.push({dist[v], v});
            }
        }
    }
}

int32_t main()
{
    cin >> N >> M;
    for(int i = 1; i <= M; i++)
    {
        int u, v, c, d;
        cin >> u >> v >> c >> d;
        ar[i] = make_tuple(u, v, c, d);
        g[u].push_back({v, c});
    }

    for(int i = 0; i <= M; i++)
    {
        int D = 0;
        if(i != 0)
        {
            auto[u, v, c, d] = ar[i];
            for(auto it = g[u].begin(); it != g[u].end(); it++)
            {
                if(it->first == v && it->second == c)
                {
                    g[u].erase(it);
                    break;
                }
            }
            g[v].push_back({u, c});
            D = d;
        }
        dijkstra(1);
        int p1 = dist[N]-1;
        dijkstra(N);
        int p2 = dist[1]-1;
        if(p1 != -1 && p2 != -1) ans = min(ans, p1+p2+D);
        if(i != 0)
        {
            auto[u, v, c, d] = ar[i];
            for(auto it = g[v].begin(); it != g[v].end(); it++)
            {
                if(it->first == u && it->second == c)
                {
                    g[v].erase(it);
                    break;
                }
            }
            g[u].push_back({v, c});
        }
    }
    // cout << (ans > comp ? -1 : ans);
    cout << ans;
    return 0;
}


// sub 1 -> ok  brute the one you'll invert
// sub 2 -> ok  after inverting you can still use the other way
// for each guy, have P(1, i) P(i, N) P(N, i) P(i, 1). For each V ans = +1, min(+2, P(U, N)+d+c), +3, min(+4, P(U,1)+d+c)
// sub 3 -> choose one and just get to the end

• Subtask 1: 0 / 5
• Subtask 2: 0 / 11
• Subtask 3: 0 / 21
• Subtask 4: 0 / 63