제출 #380225

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
380225		ignaciocanta	Commuter Pass (JOI18_commuter_pass)	C++14	0 / 100	1476 ms	262148 KiB

commuter_pass

#include <bits/stdc++.h>
//#include "crocodile.h"
 
using namespace std;
 
using tint = long long;
using ld = long double;
 
#define FOR(i,a,b) for (int i = (a); i < (b); ++i)
#define F0R(i,a) FOR(i,0,a)
#define ROF(i,a,b) for (int i = (b)-1; i >= (a); --i)
#define R0F(i,a) ROF(i,0,a)
#define trav(a,x) for (auto& a: x)
 
using pi = pair<int,int>;
using pl = pair<tint,tint>;
using vi = vector<int>;
using vpi = vector<pi>;
using vpl = vector<pl>;
using vvi = vector<vi>;
using vl = vector<tint>;
using vb = vector<bool>;
 
#define pb push_back
#define pf push_front
#define rsz resize
#define all(x) begin(x), end(x)
#define rall(x) x.rbegin(), x.rend() 
#define sz(x) (int)(x).size()
#define ins insert
 
#define f first
#define s second
#define mp make_pair
 
#define DBG(x) cerr << #x << " = " << x << endl;
 
const int MOD = 1e9+7;
const tint mod = 998244353;
const int MX = 1e5+5;
const tint INF = 1e18; 
const int inf = 2e9;
const ld PI = acos(ld(-1)); 
const ld eps = 1e-5;
 
const int dx[4] = {1, -1, 0, 0};
const int dy[4] = {0, 0, 1, -1};
 
template<class T> void remDup(vector<T> &v){ 
    sort(all(v)); v.erase(unique(all(v)),end(v));
}
 
template<class T> bool ckmin(T& a, const T& b) {
    return b < a ? a = b, 1 : 0; 
} 
template<class T> bool ckmax(T& a, const T& b) {
    return a < b ? a = b, 1 : 0; 
}
 
bool valid(int x, int y, int n, int m){
    return (0<=x && x<n && 0<=y && y<m);
}
 
int cdiv(int a, int b) { return a/b+((a^b)>0&&a%b); } //redondea p arriba
int fdiv(int a, int b) { return a/b-((a^b)<0&&a%b); } //redondea p abajo
 
void NACHO(string name = ""){
    ios_base::sync_with_stdio(0); cin.tie(0);
    if(sz(name)){
		freopen((name+".in").c_str(), "r", stdin);
		freopen((name+".out").c_str(), "w", stdout);
	}
}

vpi adj[MX];
vi dag[MX];
tint dS[MX], dT[MX], dU[MX], dV[MX], dpU[MX], dpV[MX];
bool vis[MX];
int n;

void dijkstra(int node, tint a[]){
	F0R(i, n) vis[i] = 0, a[i] = inf;
	a[node] = 0;
	priority_queue<pi, vpi, greater<pi>> q;
	q.push(mp(0, node));
	while(sz(q)){
		auto u = q.top().s; q.pop();
		if(vis[u]) continue;
		vis[u] = 1;
		trav(v, adj[u]){
			if(a[u]+v.s < a[v.f]){
				a[v.f] = a[u]+v.s;
				q.push(mp(a[v.f], v.f));
			}
		}
	}
}

vi ord;

void dfs(int node){
	trav(u, dag[node]){
		dfs(u);
	}
	ord.pb(node);
}

int main(){
	NACHO();
	// problema jodido
	// consideremos el dijkstra dag
	// nuestro camino óptimo va a consistir en: u -> x -> y -> v
	// donde x e y son nodos del dag.
	// sea dpU[i] el minimo costo en ir de u a un nodo x en el camino u->i del dag
	// sea dpV[i] el minimo costo en ir de v a un nodo x en el camino u->I del dag
	// la respuesta es dpU[v]+dpV[v]
	int m, s, t, u, v;
	cin >> n >> m >> s >> t >> u >> v;
	--s, --t, --u, --v;
	F0R(i, m){
		int a, b, c; cin >> a >> b >> c;
		--a, --b;
		adj[a].pb(mp(b, c)); adj[b].pb(mp(a, c));
	}
	dijkstra(s, dS); dijkstra(t, dT); dijkstra(u, dU); dijkstra(v, dV);
	F0R(i, n){
		dpU[i] = inf; dpV[i] = inf;
		trav(u, adj[i]){
			if(dS[i]+u.s+dT[u.f] == dT[s]) dag[i].pb(u.f);
		}
	}
	dpU[s] = dU[s]; dpV[s] = dV[s];
	dfs(s);
	reverse(all(ord));
	F0R(i, sz(ord)){
		trav(u, dag[ord[i]]){
			dpU[u] = dU[u];
			ckmin(dpU[u], dpU[ord[i]]);
			dpV[u] = dV[u];
			ckmin(dpV[u], dpV[ord[i]]);
		}
	}
	cout << min(dpU[t]+dpV[t], dU[v]) << "\n";
}

컴파일 시 표준 에러 (stderr) 메시지

commuter_pass.cpp: In function 'void NACHO(std::string)':
commuter_pass.cpp:70:10: warning: ignoring return value of 'FILE* freopen(const char*, const char*, FILE*)', declared with attribute warn_unused_result [-Wunused-result]
   70 |   freopen((name+".in").c_str(), "r", stdin);
      |   ~~~~~~~^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
commuter_pass.cpp:71:10: warning: ignoring return value of 'FILE* freopen(const char*, const char*, FILE*)', declared with attribute warn_unused_result [-Wunused-result]
   71 |   freopen((name+".out").c_str(), "w", stdout);
      |   ~~~~~~~^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

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