제출 #1228054

#제출 시각아이디문제언어결과실행 시간메모리
1228054Dzadzo사이버랜드 (APIO23_cyberland)C++20
20 / 100
1026 ms2162688 KiB
#include <bits/stdc++.h> #include "cyberland.h" using namespace std; double solve(int N, int M, int K, int H, vector<int> x, vector<int> y, vector<int> c, vector<int> arr) { // build graph 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]}); } // dp[v][d] = min time to reach v having used d divides const double INF = 1e18; vector<vector<double>> dp(N, vector<double>(K+1, INF)); dp[0][0] = 0.0; // min‑heap by (time, v, used_divs) using State = tuple<double,int,int>; priority_queue<State, vector<State>, greater<State>> pq; pq.emplace(0.0, 0, 0); while (!pq.empty()) { auto [t, v, d] = pq.top(); pq.pop(); if (t > dp[v][d]) continue; if (v == H) break; // once we pop H with current best t, that's optimal for (auto [u, w] : adj[v]) { // 1) move without using ability double t1 = t + w; if (t1 < dp[u][d]) { dp[u][d] = t1; pq.emplace(t1, u, d); } // 2) if u has reset ability if (arr[u] == 0) { double t0 = 0.0; if (t0 < dp[u][d]) { dp[u][d] = t0; pq.emplace(t0, u, d); } } // 3) if u has divide‑by‑2 and we have quota if (arr[u] == 2 && d < K) { double t2 = (t + w) / 2.0; if (t2 < dp[u][d+1]) { dp[u][d+1] = t2; pq.emplace(t2, u, d+1); } } } } // answer is the best over all ways to arrive at H double ans = INF; for (int d = 0; d <= K; d++) ans = min(ans, dp[H][d]); return ans == INF ? -1.0 : ans; }
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