제출 #750010

#제출 시각아이디문제언어결과실행 시간메모리
750010Cookie사이버랜드 (APIO23_cyberland)C++17
100 / 100
1996 ms128952 KiB
#include<bits/stdc++.h>
#include "cyberland.h"
#include<fstream>
using namespace std;
ifstream fin("VNOICUP.INP");
ofstream fout("VNOICUP.OUT");
#define sz(a) (int)a.size()
#define ll long long
#define pb push_back
#define forr(i, a, b) for(int i = a; i < b; i++)
#define dorr(i, a, b) for(int i = a; i >= b; i--)
#define ld long double
#define vt vector
#include<fstream>
#define fi first
#define se second
#define pll pair<ll, ll>
#define pii pair<int, int>
const int mxn = 1e5;
const ll prec = 1e-6;
struct ch{
    int u, take; ld dd;
};
struct cmp{
    bool operator()(const ch &a, const ch &b){
        return(a.dd > b.dd);
    }
};
vt<pii>adj[mxn + 1];
ld d[mxn + 1][71];
bool vis[mxn + 1], go[mxn + 1][71];
void dfs(int s){
    vis[s] = 1;
    for(auto [v, w]: adj[s]){
        if(!vis[v])dfs(v);
    }
}
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) {
    k = min(k, 70);
    forr(i, 0, n){
        adj[i].clear();
        forr(j, 0, k + 1){
            d[i][j] = 1e18; go[i][j] = 0;
        }
        vis[i] = 0;
    }
    forr(i, 0, m){
        int u = x[i], v = y[i], w = c[i];
        adj[u].pb({v, w}); adj[v].pb({u, w});
    }
    dfs(0);
    if(!vis[h])return(-1);
    
    d[0][0] = 0;
    for(int tt = 0; tt <= k; tt++){
        priority_queue<ch, vt<ch>, cmp>pq; 
        for(int i = 0; i < n; i++){
            if(d[i][tt] != 1e18)pq.push({i, tt, d[i][tt]});
        }
    while(!pq.empty()){
        auto [u, take, dd] = pq.top(); pq.pop();
        if((abs(dd - d[u][take]) > prec) || u == h)continue;
        for(auto [v, w]: adj[u]){
            if(arr[v] == 0 && d[v][tt] != 0){
                d[v][tt] = 0;
                pq.push({v, tt, d[v][tt]});
                continue;
            }
            if(d[v][take] > d[u][take] + w){
                d[v][take] = d[u][take] + w;
                pq.push({v, take, d[v][take]});
            }
            if(arr[u] == 2 && take < k){
                d[v][take + 1] = min(d[v][take + 1], (d[u][take] / 2.0) + w);
            }
        }
    }
    }
    ld ans = 1e18;
    for(int i = 0; i <= k; i++)ans = min(ans, d[h][i]);
    return(ans);
}
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