제출 #892445

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
892445		d4xn	사이버랜드 (APIO23_cyberland)	C++17	44 / 100	33 ms	12112 KiB

cyberland

#include "cyberland.h"
#include <bits/stdc++.h>
using namespace std;

#define ll long long
#define ld long double

const int N = 1e5+5;

int n, h;
ll mn[N], dis[3][N];
vector<int> arr;
vector<pair<int, ll>> adj[N];

void dijkstra(int x, int idx) {
    priority_queue<pair<ll, int>> pq;
    if (idx == 2) {
      for (int i = 0; i < n; i++) {
        if (arr[i] || dis[0][i] == -1) continue;
        pq.push(make_pair(-0, i));
      }
    }
    else {
      pq.push(make_pair(-0, x));
    }

    while (!pq.empty()) {
      ll d = -pq.top().first;
      int y = pq.top().second;
      pq.pop();

      if (dis[idx][y] != -1) continue;
      dis[idx][y] = d;

      if (idx != 1 && y == h) continue;
      for (auto& [z, d2] : adj[y]) {
        if (dis[idx][z] != -1) continue;
        pq.push(make_pair(-(d+d2), z));
      }
    }
}

double solve(int N, int m, int k, int H, vector<int> x, vector<int> y, vector<int> c, vector<int> ARR) {
    n = N;
    h = H;
    arr.clear();
    arr = ARR;
    for (int i = 0; i < n; i++) {
      adj[i].clear();
    }
    
    for (int i = 0; i < m; i++) {
        adj[x[i]].push_back(make_pair(y[i], c[i]));
        adj[y[i]].push_back(make_pair(x[i], c[i]));
    }

    for (int i = 0; i < n; i++) {
        dis[0][i] = dis[1][i] = dis[2][i] = -1;

        mn[i] = -1;
        for (auto& [j, d] : adj[i]) {
          if (j == h) continue;
          
          if (mn[i] == -1) mn[i] = d;
          else mn[i] = min(mn[i], d);
        }
    }
    dijkstra(0, 0);
    dijkstra(h, 1);
    dijkstra(-1, 2);

    // nada
    ld a = dis[0][h];
    ld ans = a;
    
    // solo 0
    a = dis[2][h];
    if (dis[2][h] != -1) ans = min(ans, a);

    // solo 2
    for (int i = 0; i < n; i++) {
      if (arr[i] != 2 || dis[0][i] == -1 || dis[1][i] == -1) continue;

      ld div = 2;
      a = dis[0][i] / div;

      ld b = dis[1][i];
      ans = min(ans, a + b);

      if (mn[i] != -1 && k > 1) {
        int l = 1;
        int r = k-1;
        while (l < r) {
          ld mid = (l+r)/2;

          ld val1 = a / (div*mid) + mn[i]*(mid-1) / div;
          ld val2 = a / (div*(mid+1)) + mn[i]*(mid) / div;

          if (val1 < val2) r = mid;
          else l = mid+1;
        }
        ld nwA = a / (div*l) + mn[i]*(l-1) / div;
        ans = min(ans, nwA + b);
      }
    }
    
    // 0 y 2
    for (int i = 0; i < n; i++) {
      if (arr[i] != 2 || dis[0][i] == -1 || dis[1][i] == -1 || dis[2][i] == -1) continue;

      ld div = 2;
      a = dis[2][i] / div;

      ld b = dis[1][i];
      ans = min(ans, a + b);

      if (mn[i] != -1 && k > 1) {
        int l = 1;
        int r = k-1;
        while (l < r) {
          ld mid = (l+r)/2;

          ld val1 = a / (div*mid) + mn[i]*(mid-1) / div;
          ld val2 = a / (div*(mid+1)) + mn[i]*(mid) / div;

          if (val1 < val2) r = mid;
          else l = mid+1;
        }
        ld nwA = a / (div*l) + mn[i]*(l-1) / div;
        ans = min(ans, nwA + b);
      }
    }
    return ans;
}

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