結果

問題 No.1449 新プロランド
ユーザー leaf_1415leaf_1415
提出日時 2021-03-31 15:01:08
言語 C++11
(gcc 11.4.0)
結果
RE  
実行時間 -
コード長 5,404 bytes
コンパイル時間 907 ms
コンパイル使用メモリ 92,584 KB
実行使用メモリ 135,360 KB
最終ジャッジ日時 2024-06-06 10:55:16
合計ジャッジ時間 23,135 ms
ジャッジサーバーID
（参考情報） 		judge5 / judge2
このコードへのチャレンジ
（要ログイン）

テストケース

テストケース表示
入力 結果 実行時間
実行使用メモリ
testcase_00 RE -
testcase_01 RE -
testcase_02 RE -
testcase_03 RE -
testcase_04 RE -
testcase_05 RE -
testcase_06 RE -
testcase_07 RE -
testcase_08 RE -
testcase_09 RE -
testcase_10 RE -
testcase_11 RE -
testcase_12 RE -
testcase_13 RE -
testcase_14 RE -
testcase_15 RE -
testcase_16 RE -
testcase_17 RE -
testcase_18 RE -
testcase_19 RE -
testcase_20 RE -
testcase_21 RE -
testcase_22 RE -
testcase_23 RE -
testcase_24 RE -
testcase_25 RE -
testcase_26 RE -
testcase_27 RE -
testcase_28 RE -
権限があれば一括ダウンロードができます

ソースコード

diff # 

#include <iostream>
#include <cstdio>
#include <cmath>
#include <ctime>
#include <cstdlib>
#include <cassert>
#include <vector>
#include <list>
#include <stack>
#include <queue>
#include <deque>
#include <map>
#include <set>
#include <bitset>
#include <string>
#include <algorithm>
#include <utility>
#include <complex>
#define rep(x, s, t) for(llint (x) = (s); (x) <= (t); (x)++)
#define reps(x, s) for(llint (x) = 0; (x) < (llint)(s).size(); (x)++)
#define chmin(x, y) (x) = min((x), (y))
#define chmax(x, y) (x) = max((x), (y))
#define sz(x) ((ll)(x).size())
#define ceil(x, y) (((x)+(y)-1) / (y))
#define all(x) (x).begin(),(x).end()
#define outl(...) dump_func(__VA_ARGS__)
#define inf 1e18

using namespace std;

typedef long long llint;
typedef long long ll;
typedef pair<ll, ll> P;

struct edge{
	ll to, cost;
	edge(){}
	edge(ll a, ll b){ to = a, cost = b;}
};
const ll dx[] = {1, 0, -1, 0}, dy[] = {0, -1, 0, 1};

const ll mod = 1000000007;
//const ll mod = 998244353;

struct mint{
	ll x = 0;
	mint(ll y = 0){x = y; if(x < 0 || x >= mod) x = (x%mod+mod)%mod;}
	mint(const mint &ope) {x = ope.x;}

	mint operator-(){return mint(-x);}
	mint operator+(const mint &ope){return mint(x) += ope;}
	mint operator-(const mint &ope){return mint(x) -= ope;}
	mint operator*(const mint &ope){return mint(x) *= ope;}
	mint operator/(const mint &ope){return mint(x) /= ope;}
	mint& operator+=(const mint &ope){
		x += ope.x;
		if(x >= mod) x -= mod;
		return *this;
	}
	mint& operator-=(const mint &ope){
		x += mod - ope.x;
		if(x >= mod) x -= mod;
		return *this;
	}
	mint& operator*=(const mint &ope){
		x *= ope.x, x %= mod;
		return *this;
	}
	mint& operator/=(const mint &ope){
		ll n = mod-2; mint mul = ope;
		while(n){
			if(n & 1) *this *= mul;
			mul *= mul;
			n >>= 1;
		}
		return *this;
	}
	mint inverse(){return mint(1) / *this;}
	bool operator ==(const mint &ope){return x == ope.x;}
	bool operator !=(const mint &ope){return x != ope.x;}
};
mint modpow(mint a, ll n){
	if(n == 0) return mint(1);
	if(n % 2) return a * modpow(a, n-1);
	else return modpow(a*a, n/2);
}
istream& operator >>(istream &is, mint &ope){
	ll t; is >> t, ope.x = t;
	return is;
}
ostream& operator <<(ostream &os, mint &ope){return os << ope.x;}
ostream& operator <<(ostream &os, const mint &ope){return os << ope.x;}

bool exceed(ll x, ll y, ll m){return x >= m / y + 1;}
void mark(){ cout << "*" << endl; }
void yes(){ cout << "Yes" << endl; }
void no(){ cout << "No" << endl; }
ll gcd(ll a, ll b){if(b == 0) return a; return gcd(b, a%b);}
ll digitnum(ll x, ll b = 10){ll ret = 0; for(; x; x /= b) ret++; return ret;}
ll digitsum(ll x, ll b = 10){ll ret = 0; for(; x; x /= b) ret += x % b; return ret;}

template<typename T>
ostream& operator << (ostream& os, vector<T>& vec) {
	for(int i = 0; i < vec.size(); i++) {
		os << vec[i] << (i + 1 == vec.size() ? "" : " ");
	}
	return os;
}
template<typename T>
ostream& operator << (ostream& os, deque<T>& deq) {
	for(int i = 0; i < deq.size(); i++) {
		os << deq[i] << (i + 1 == deq.size() ? "" : " ");
	}
	return os;
}
template<typename T, typename U>
ostream& operator << (ostream& os, pair<T, U>& pair_var) {
	os << "(" << pair_var.first << ", " << pair_var.second << ")";
	return os;
}
template<typename T, typename U>
ostream& operator << (ostream& os, const pair<T, U>& pair_var) {
	os << "(" << pair_var.first << ", " << pair_var.second << ")";
	return os;
}
template<typename T, typename U>
ostream& operator << (ostream& os, map<T, U>& map_var) {
	for(typename map<T, U>::iterator itr = map_var.begin(); itr != map_var.end(); itr++) {
		os << "(" << itr->first << ", " << itr->second << ")";
		itr++;
		if(itr != map_var.end()) os << ",";
		itr--;
	}
	return os;
}
template<typename T>
ostream& operator << (ostream& os, set<T>& set_var) {
	for(typename set<T>::iterator itr = set_var.begin(); itr != set_var.end(); itr++) {
		os << *itr;
		++itr;
		if(itr != set_var.end()) os << " ";
		itr--;
	}
	return os;
}
template<typename T>
void outa(T a[], ll s, ll t)
{
	for(ll i = s; i <= t; i++){ cout << a[i]; if(i < t) cout << " ";}
	cout << endl;
}
void dump_func() {cout << endl;}
template <class Head, class... Tail>
void dump_func(Head &&head, Tail &&... tail) {
	cout << head;
	if(sizeof...(Tail) > 0) cout << " ";
	dump_func(std::move(tail)...);
}

void dijkstra(vector<edge> G[], llint S, llint dist[])
{
	for(int i = 0; i <= 2000000; i++) dist[i] = inf;
	dist[S] = 0;

	priority_queue< P, vector<P>, greater<P> > Q;
	Q.push( make_pair(0, S) );

	llint v, d;
	while(Q.size()){
		d = Q.top().first;
		v = Q.top().second;
		Q.pop();
		if(dist[v] < d) continue;
		for(int i = 0; i < G[v].size(); i++){
			if(dist[G[v][i].to] > d + G[v][i].cost){
				dist[G[v][i].to] = d + G[v][i].cost;
				Q.push( make_pair(dist[G[v][i].to], G[v][i].to) );
			}
		}
	}
}

ll n, m;
ll u[105], v[105], w[105];
vector<edge> G[2000000];
ll dist[2000000];
ll t[105];

int main(void)
{
	ios::sync_with_stdio(0);
	cin.tie(0);

	ll T = 15000;

	cin >> n >> m;
	rep(i, 1, m) cin >> u[i] >> v[i] >> w[i];
	rep(i, 1, n) cin >> t[i];

	rep(i, 1, m){
		rep(j, 0, T){
			if(j+t[u[i]] && j+t[u[i]] <= T) G[j*n+u[i]].push_back(edge((j+t[u[i]])*n+v[i], w[i]/(j+t[u[i]])));
			if(j+t[v[i]] && j+t[v[i]] <= T) G[j*n+v[i]].push_back(edge((j+t[v[i]])*n+u[i], w[i]/(j+t[v[i]])));
		}
	}
	dijkstra(G, 1, dist);

	ll ans = inf;
	rep(i, 0, T) chmin(ans, dist[i*n+n]+i);
	outl(ans);

	return 0;
}
0