Submission #1097657

#	Time	Username	Problem	Language	Result	Execution time	Memory
1097657		KindaGoodGames	Robot (JOI21_ho_t4)	C++14	0 / 100	283 ms	30524 KiB

Main

#include<bits/stdc++.h>

#define ll long long
#define int ll
#define pii pair<int,int>
#define tiii tuple<int,int,int>
#define tiiii tuple<int,int,bool,int>

using namespace std;

int INF = numeric_limits<ll>::max()/2;
int32_t main(){
    int n, m;
    cin >> n >> m;
    vector<vector<tiii>> adj(n);
    vector<int> colorAmt(m);
    for(int i = 0; i < m; i++){
        int a, b, c, v;
        cin >> a >> b >> c >> v;
        c--;
        a--;b--;
        adj[a].push_back({b,c,v});
        adj[b].push_back({a,c,v});

        colorAmt[c]++;
    }

    // Dijkstra

    // We use the following Idea: We can either change the current edge or all other edges. If we change all other edges, we need to keep track
    // of which ones we changed
 
    vector<int> dist(n, INF);

    // distance | vertex | color | cost | precessor
    priority_queue<tiiii, vector<tiiii>, greater<tiiii>> pq;
    pq.push({0,0,0,0});
    while(pq.size()){
        int d, v, c , pre;
        tie(d,v,c, pre) = pq.top(); pq.pop();
        if(d >= dist[v]) continue;
        dist[v] = d; 

        // get all costs
        map<int,int> memo; 
        for(auto e : adj[v]){
            int u, co, we;
            tie(u,co,we) = e;
            // if the colors don't match or it's the edge backwards, ignore
            // || get<0>(e) == get<0>(p) 
            memo[get<1>(e)] += we;
        }
        for(auto p : adj[v]){
            if(get<0>(p) == pre) continue;
            int s = memo[get<1>(p)]; 

            // only taking one edge
            if(get<2>(p) < s-get<2>(p)){
                pq.push({get<2>(p)+d, get<0>(p), 0, v});
            }else{
                pq.push({s+d-get<2>(p), get<0>(p), 1, v});
            }
        }
        // either change the color of this edge or all others
    }
    if(dist[n-1]>=INF/2) dist[n-1] = -1; 
    cout << dist[n-1] << endl;
}

• Subtask 1: 0 / 34
• Subtask 2: 0 / 24
• Subtask 3: 0 / 42