제출 #632644

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
632644		Karliver	Olympic Bus (JOI20_ho_t4)	C++17	0 / 100	27 ms	2828 KiB

ho_t4

#pragma GCC optimize ("O3")
#pragma GCC target ("sse4")
#include <bits/stdc++.h>

#define FIXED_FLOAT(x)  std::fixed <<std::setprecision(20) << (x)
#define all(v) (v).begin(), (v).end()
using namespace  std;
#define forn(i,n) for (int i = 0; i < (n); ++i)
#define rforn(i, n) for(int i = (n) - 1;i >= 0;--i)
#define sz(x) (int)x.size()
#define ff first
#define se second
#define mp make_pair
using ll = long long;
int mod = (ll)1e9 + 7;
const int INF = 1e9 + 1;
const int N = 205;
const double eps = 1e-7;

template <class T> using V = vector<T>;  
template <class T> using VV = V<V<T>>;  
template<class T, size_t SZ> using AR = array<T, SZ>;
template<class T> using PR = pair<T, T>;
template <typename XPAX>
bool ckma(XPAX &x, XPAX y) {
    return (x < y ? x = y, 1 : 0);
}
template <typename XPAX>
bool ckmi(XPAX &x, XPAX y) {
    return (x > y ? x = y, 1 : 0);
}



void __print(int x) {cerr << x;}
void __print(long x) {cerr << x;}
void __print(long long x) {cerr << x;}
void __print(unsigned x) {cerr << x;}
void __print(unsigned long x) {cerr << x;}
void __print(unsigned long long x) {cerr << x;}
void __print(float x) {cerr << x;}
void __print(double x) {cerr << x;}
void __print(long double x) {cerr << x;}
void __print(char x) {cerr << '\'' << x << '\'';}
void __print(const char *x) {cerr << '\"' << x << '\"';}
void __print(const string &x) {cerr << '\"' << x << '\"';}
void __print(bool x) {cerr << (x ? "true" : "false");}

template<typename T, typename V>
void __print(const pair<T, V> &x) {cerr << '{'; __print(x.first); cerr << ','; __print(x.second); cerr << '}';}
template<typename T>
void __print(const T &x) {int f = 0; cerr << '{'; for (auto &i: x) cerr << (f++ ? "," : ""), __print(i); cerr << "}";}
void _print() {cerr << "]\n";}
template <typename T, typename... V>
void _print(T t, V... v) {__print(t); if (sizeof...(v)) cerr << ", "; _print(v...);}
#define debug(x...) cerr << "[" << #x << "] = ["; _print(x)
const int  MX = 5e5 + 10;
const ll inf = 1e17;
int A[MX], B[MX], C[MX], D[MX];
int n, m;
vector<int> g[N][2];
void dij(int fr, int tr, int tp, vector<ll>& d, vector<bool> &vis) {
	d.resize(n + 1, inf);
	vis.resize(m + 1, false);
	vector<bool> seen(n + 1, false);
	vector<int> par(n + 1, -1);
	d[fr] = 0;
	while(1) {
		int v = n;
		forn(i, n) if(!seen[i]&& d[i]<d[v])v = i;
		if(v == n)break;
		seen[v] = true;
		for(int i : g[v][tp]) {
			int y = A[i] ^ B[i] ^ v;
			if(d[v]+C[i] < d[y]) {
				d[y] = d[v]+C[i];
				par[y] = i;
			}
		}
	}
	if(tp)return;
	int y = tr;
	while(par[y] != -1) {
		vis[par[y]] = true;
		y = A[par[y]];
	}
}

ll without(int fr, int tr, int  cant) {
	
	vector<ll> d(n + 1, inf);
	vector<bool> vis(n + 1, false);
	d[fr] = 0;
	
	while(1) {
		int v = n;
		forn(i, n) if(!vis[i]&&d[i]<d[v])v = i;
		if(v == n)break;
		vis[v] = true;
		for(int i : g[v][0]) if(i != cant) {
			int y = A[i] ^ B[i] ^ v;
			if(d[v]+C[i] < d[y]) {
				d[y] = d[v]+C[i];
			}
		}
		if(B[cant] == v) {
			ckmi(d[A[cant]], d[v]+C[cant]);
		}
	}
	return d[tr];
}
		
void solve() {	
	cin >> n >> m;
	
	for(int i = 0;i < m;++i) {
		cin >> A[i] >> B[i] >> C[i] >> D[i];
		--A[i];
		--B[i];
		g[A[i]][0].push_back(i);
		g[B[i]][1].push_back(i);
	}
	vector<bool> vis1, vis2, tmp;
	vector<ll> d1, d2, d3, d4;
	
	dij(0, n - 1,0, d1, vis1);

	dij(n - 1, 0, 0, d2, vis2);
	dij(0, n - 1, 1, d3, tmp);
	dij(n - 1, 0,1,  d4, tmp);
	ll res = inf;
	
	for(int i = 0;i < m;++i) {
		ll cost = D[i];
		
		if(vis1[i])cost += without(0, n - 1, i);
		else cost += min(d1[n - 1], d1[B[i]]+C[i]+d4[A[i]]);
		
		if(vis2[i]) cost += without(n - 1, 0, i);
		else cost += min(d2[0], d2[B[i]]+C[i]+d3[A[i]]);
		//debug(i, cost);
		ckmi(res, cost);
	}
	
	cout << (res == inf ? -1 : res) << '\n';
		
		
}
void test_case() {
    int t;
    cin >> t;
    forn(p, t) {

        //cout << "Case #" << p + 1 << ": ";
        solve();
    }
}
int main() {

    ios::sync_with_stdio(false);
    cin.tie(0);

    solve();

}
         
    

• Subtask 1: 0 / 5
• Subtask 2: 0 / 11
• Subtask 3: 0 / 21
• Subtask 4: 0 / 63