Submission #751014

#	Time	Username	Problem	Language	Result	Execution time	Memory
751014		Gurban	Cyberland (APIO23_cyberland)	C++17	97 / 100	3070 ms	11872 KiB

cyberland

#include "cyberland.h"
#include <bits/stdc++.h>

using namespace std;

using ll = long long;
using ld = long double;

const double inf = 1e16;
const int maxn=1e5+5;
const double eps = 0.00000001;
int D[maxn];
vector<pair<int,double>>E[maxn];

int p[maxn];

int ata(int x){
    if(p[x] == x) return x;
    return p[x] = ata(p[x]);
}

void un(int x,int y){
    int aa = ata(x);
    int bb = ata(y);
    p[bb] = aa;
}

bool con(int x,int y){
    int aa = ata(x);
    int bb = ata(y);
    return (aa == bb);
}

double solve(int N, int M, int K, int H, vector<int> x, vector<int> y, vector<int> c, vector<int> arr) {
    for(int i = 0;i < N;i++) E[i].clear(),p[i] = i;
    for(int i = 0;i < M;i++){
        E[x[i]].push_back({y[i],c[i]});
        E[y[i]].push_back({x[i],c[i]});
        if(x[i] != H and y[i] != H) un(x[i],y[i]);
    }
    vector<int>zeros = {0};
    vector<int>two;
    for(int i = 1;i < N;i++) if(arr[i] == 0 and con(0,i)) zeros.push_back(i);
    for(int i = 1;i < N;i++) if(arr[i] == 2) two.push_back(i);
    
    priority_queue<pair<double,int>>q;
    q.push({0,H});

    double ans = inf;
    double der = 1;

    for(int i = 0;i <= K;i++){
        vector<double>dis(N,inf);
        vector<bool>vis(N,0);
        while(!q.empty()){
            
            int x = q.top().second;
            double len = -q.top().first;
            q.pop();

            if(vis[x]) continue;
            vis[x] = 1;

            for(auto j : E[x]){
                if(j.first == H) continue;
                double now = (j.second / der) + len;
                if(dis[j.first] > now){
                    dis[j.first] = now;
                    q.push({-now,j.first}); 
                }
            }
        }
        for(auto j : zeros) ans = min(ans,dis[j]);
        for(auto j : two) if(abs(dis[j] - inf) >= eps) q.push({-dis[j],j});
        der *= 2;
    }
    if(abs(ans - inf) < eps) return -1;
    return ans;
}

• 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: 29 / 29
• Subtask 8: 0 / 3