Submission #1253828

#	Time	Username	Problem	Language	Result	Execution time	Memory
1253828		anfi	Cyberland (APIO23_cyberland)	C++20	15 / 100	668 ms	25444 KiB

cyberland

#include <bits/stdc++.h>
using namespace std;
#define fi first
#define se second
const double INF = 1e30;

double solve(int N, int M, int K, int H,
             vector<int> x, vector<int> y, vector<int> c, vector<int> arr) {
    K = min(K, 100);
    vector<vector<pair<int,int>>> adj(N);
    for(int i = 0; i < M; i++){
        adj[x[i]].push_back({y[i], c[i]});
        adj[y[i]].push_back({x[i], c[i]});
    }

    vector dp(N, vector<double>(K+1, INF));
    priority_queue< tuple<double,int,int>,
                    vector<tuple<double,int,int>>,
                    greater<tuple<double,int,int>> > pq;

    dp[0][0] = 0;
    pq.push({0, 0, 0});
    while(!pq.empty()){
        auto [dist, u, used] = pq.top(); pq.pop();
        if(dist > dp[u][used]) continue;
        for(auto [v, w] : adj[u]){
            double nd = dist + w;
            if(nd < dp[v][used]){
                dp[v][used] = nd;
                pq.push({nd, v, used});
            }
        }
    }

    for(int i = 0; i < N; i++){
        if(arr[i] == 0 && dp[i][0] > 0){
            dp[i][0] = 0;
            pq.push({0, i, 0});
        }
    }

    while(!pq.empty()){
        auto [dist, u, used] = pq.top(); pq.pop();
        if(dist > dp[u][used]) continue;

        for(auto [v, w] : adj[u]){
            if(dist + w < dp[v][used]){
                dp[v][used] = dist + w;
                pq.push({dp[v][used], v, used});
            }
            if(arr[u] == 2 && used < K){
                double cd = dist + w/2.0;
                if(cd < dp[v][used+1]){
                    dp[v][used+1] = cd;
                    pq.push({cd, v, used+1});
                }
            }
        }

        if(u == H && arr[H]==2 && used < K){
            if(dist/2.0 < dp[H][used+1]){
                dp[H][used+1] = dist/2.0;
                pq.push({dp[H][used+1], H, used+1});
            }
        }
    }

    double ans = INF;
    for(int d = 0; d <= K; d++) ans = min(ans, dp[H][d]);
    return (ans >= INF/2 ? -1.0 : ans);
}

• Subtask 1: 0 / 5
• Subtask 2: 8 / 8
• Subtask 3: 0 / 13
• Subtask 4: 0 / 19
• Subtask 5: 7 / 7
• Subtask 6: 0 / 16
• Subtask 7: 0 / 29
• Subtask 8: 0 / 3