Submission #1228081

#	Time	Username	Problem	Language	Result	Execution time	Memory
1228081		pera	Cyberland (APIO23_cyberland)	C++20	68 / 100	3095 ms	89156 KiB

cyberland

#include "cyberland.h"
#include <bits/stdc++.h>
#define LL __int128
using namespace std;
const int N = 1E5 + 1 , W = 65;
LL dist[N][W] , P[W];
vector<pair<int , LL>> g[N];
double solve(int N, int M, int K, int H, std::vector<int> x, std::vector<int> y, std::vector<int> c, std::vector<int> arr) {
   for(int i = 0;i < N;i ++){
      g[i].clear();
   }
   P[0] = 1;
   for(int i = 1;i < W;i ++){
      P[i] = P[i - 1] * (LL)2;
   }
   for(int i = 0;i < N;i ++){
      for(int j = 0;j <= K;j ++){
         dist[i][j] = -1;
      }
   }
   for(int i = 0;i < M;i ++){
      g[x[i]].emplace_back(y[i] , c[i]);
      g[y[i]].emplace_back(x[i] , c[i]);
   }
   K = min(K , 64);
   priority_queue<pair<LL , pair<int , int>>> pq;
   dist[0][0] = 0;
   LL ans = -1;
   int p = 0;
   pq.push(pair<LL , pair<int , int>>{0 , pair<int , int>{0 , 0}});
   while(!pq.empty()){
      auto [D , V] = pq.top();
      auto [u , x] = V;
      pq.pop();
      if(dist[u][x] != D){
         continue;
      }
      if(u == H){
         if(ans == -1){
            ans = dist[H][x];
            p = x;
            continue;
         }
         LL left = ans * P[x];
         LL right = dist[H][x] * P[p];
         if(right < left){
            ans = dist[H][x];
            p = x;
         }
         continue;
      }
      for(auto [v , w] : g[u]){
         LL nD = D + w * P[x];
         if(arr[v] == 0){
            nD = 0;
         }
         if(dist[v][x] == -1 || dist[v][x] > nD){
            dist[v][x] = nD;
            pq.push(pair<LL , pair<int , int>>{dist[v][x] , {v , x}});
         }
         if(arr[v] == 2 && x < K && (dist[v][x + 1] == -1 || dist[v][x + 1] > nD)){
            dist[v][x + 1] = nD;
            pq.push(pair<LL , pair<int , int>>{dist[v][x + 1] , {v , x + 1}});
         }
      }
   }
   if(ans == -1){
      return ans;
   }
   long double S = 1;
   for(int i = 0;i < p;i ++){
      S /= 2.0;
   }
   S *= (long double)ans;
   return (double)S;
}

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