제출 #1097653

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1097653		KindaGoodGames	Robot (JOI21_ho_t4)	C++14	0 / 100	3115 ms	1247472 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<bitset<100000>> changed(n); 
    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;
        changed[v] = changed[pre];
        if(c) changed[v][pre] = c;

        for(auto p : adj[v]){
            if(get<0>(p) == pre) continue;
            int s = 0;
            // get the sum of taking all other edges
            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
                if(co != get<1>(p) || get<0>(e) == get<0>(p)) continue; 
                s += we;
            }

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

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