Submission #766308

#	Time	Username	Problem	Language	Result	Execution time	Memory
766308		OrazB	Commuter Pass (JOI18_commuter_pass)	C++17	0 / 100	366 ms	22368 KiB

commuter_pass

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <functional>
using namespace __gnu_pbds;
using namespace std;
typedef tree<int, null_type, less<int>, rb_tree_tag,
             tree_order_statistics_node_update>
    ordered_set;
//Dijkstra->set
//set.find_by_order(x) x-position value
//set.order_of_key(x) number of strictly less elements don't need *set.??
#define N 100005
#define wr cout << "Continue debugging\n";
#define all(x) (x).begin(), (x).end()
#define ll long long int
#define pii pair <ll, ll>
#define pb push_back
#define ff first
#define ss second

int n, m, s, t, u, v;
vector<pii> E[N];

void dijk(int x, vector<ll>&dis){
	set<pii> s = {{0, x}};
	dis[x] = 0;
	while(!s.empty()){
		int x = (*s.begin()).ss;
		s.erase(s.begin());
		for (auto i : E[x]){
			if (dis[i.ff] > dis[x]+i.ss){
				s.erase({dis[i.ff], i.ff});
				dis[i.ff] = dis[x]+i.ss;
				s.insert({dis[i.ff], i.ff});
			}
		}
	}
}

int main ()
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	cin >> n >> m >> s >> t >> u >> v;
	while(m--){
		int x, y, w;
		cin >> x >> y >> w;
		E[x].pb({y, w});
		E[y].pb({x, w});
	}
	vector<ll> disu(n+1, 1e18), disv(n+1, 1e18), diss(n+1, 1e18);
	dijk(v, disv); dijk(u, disu);
	vector<ll> dis(n+1, 1e18);
	vector<bool> vis(n+1, 0);
	vector<ll> canu(n+1, 1e18), canv(n+1, 1e18);
	set<pair<pii,ll>> st = {{{0, s}, 0}};
	dis[s] = 0;
	while(!st.empty()){
		int x = (*st.begin()).ff.ss, pr = (*st.begin()).ss;
		st.erase(st.begin());
		if (min(disv[x], canv[pr])+min(disu[x], canu[pr]) < canu[x]+canv[x]){
			canv[x] = min(disv[x], canv[pr]);
			canu[x] = min(disu[x], canu[pr]);
		}
		if (vis[x]) continue;
		vis[x] = 1;
		for (auto i : E[x]){
			if (dis[i.ff] >= dis[x]+i.ss){
				st.erase({{dis[i.ff], i.ff}, x});
				dis[i.ff] = dis[x]+i.ss;
				st.insert({{dis[i.ff], i.ff}, x});
			}
		}
	}
	cout << min(canu[t]+canv[t], disv[u]);
}		

• Subtask 1: 0 / 16
• Subtask 2: 0 / 15
• Subtask 3: 0 / 24
• Subtask 4: 0 / 45