이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#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) {
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) {
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;
}
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