제출 #1253826

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1253826		anfi	사이버랜드 (APIO23_cyberland)	C++20	15 / 100	529 ms	25780 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){
    double ans = 1e30;
    K = min(K, 100);
    vector<pair<int,int>> adj[N];
    vector<vector<double>> dp(N, vector<double>(K+1, inf));
    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<int> a = arr;
    priority_queue<tuple<double,int,int>, vector<tuple<double, int, int>>, greater<tuple<double,int,int>>> pq;
    dp[0][0] = 0.0;
    pq.push({0.0, 0,0});
    while(!pq.empty()){
        auto [c, u, pos] = pq.top(); pq.pop();
        if(c > dp[u][pos]) continue;
        for(auto &[v, w] : adj[u]){
            double cost = c+w;
            if(dp[v][pos] > cost){
                dp[v][pos] = cost;
                pq.push({cost, v, pos});
            }        
        }
    }
    for(int i = 0; i < N; i++){
        if(a[i] == 0 && dp[i][0] > 0){
            dp[i][0] = 0;
            pq.push({0.0, i, 0});
        }
    }
    if (dp[0][0] == 0.0) {
        pq.push({0.0, 0, 0});
    }
    while(!pq.empty()){
        auto [c, u, pos] = pq.top(); pq.pop();
        if(c > dp[u][pos]) continue;
        for(auto &[v, w] : adj[u]){
            double cost = c+w;
            if(dp[v][pos] > cost){
                dp[v][pos] = cost;
                pq.push({cost, v, pos});
            }
            if(a[u] == 2 && pos < K){
                double co = c+w/2.0;
                if(dp[v][pos+1] > co){
                    dp[v][pos+1] = co;
                    pq.push({co, v, pos+1});
                }
            }            
        }
        if(u == H && a[H] == 2 && pos < K){
            if(dp[H][pos+1] > c/2.0){
                dp[H][pos+1] = c/2.0;
                pq.push({dp[H][pos+1], H, pos+1});
            }
        }
    }
    for(int i = 0; i <= K; i++) ans = min(ans, dp[H][i]);
    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