| # | Submission time | Handle | Problem | Language | Result | Execution time | Memory |
|---|---|---|---|---|---|---|---|
| 322390 | 2020-11-14T14:49:05 Z | Farrius | Crocodile's Underground City (IOI11_crocodile) | C++11 | 0 ms | 0 KB |
#include "crocodile.h"
#include <bits/stdc++.h>
#define len(x) (int) x.size()
#define f first
#define s second
using namespace std;
using ll = long long;
using ld = long double;
const int INF = 1e9;
int travel_plan (int n, int m, int r[][2], int l[], int k, int p[]) {
//make graph
vector<pair<int, int>> g[n];
for (int i = 0; i < m; i++) {
int u = r[i][0], v = r[i][1];
g[u].push_back(make_pair(v, l[i]));
g[v].push_back(make_pair(u, l[i]));
}
//know how many exits have each vertex
set<pair<int, int>> can_exit[n];
for (int i = 0; i < k; i++) {
int exit = p[i];
for (auto u : g[exit]) {
can_exit[u.first].insert(make_pair(u.second, exit));
}
}
//dijkstra
int sol = 1e9;
vector<int> dist(n, INF);
priority_queue<pair<int, int>> pq;
bool vis[n];
memset(vis, false, sizeof(vis));
pq.push(make_pair(0, 0));
dist[0] = 0;
while (!pq.empty()) {
int v = pq.top().s;
int cur = pq.top().f;
pq.pop();
if (vis[v]) continue;
vis[v] = true;
if (len(can_exit[v]) >= 2) {
auto last = can_exit[v].rbegin();
last--;
pair<int, int> cost = *last
sol = min(sol, cost.f + cur);
continue;
}
for (auto u : g[v]) {
if (dist[u.f] > dist[v] + u.s) {
dist[u.f] = dist[v] + u.s;
pq.push(make_pair(-dist[u.f], u.f));
}
}
}
return sol;
}