제출 #480710

#제출 시각아이디문제언어결과실행 시간메모리
480710hidden1Toll (BOI17_toll)C++14
8 / 100
3084 ms14660 KiB
#include <bits/stdc++.h>
using namespace std;
//#pragma GCC optimize ("O3")
//#pragma GCC target ("sse4")
#define endl "\n"
typedef long long ll;
template<class T, class T2> inline ostream &operator <<(ostream &out, const pair<T, T2> &x) { out << x.first << " " << x.second; return out;}
template<class T, class T2> inline istream &operator >>(istream &in, pair<T, T2> &x) { in >> x.first >> x.second; return in;}
template<class T, class T2> inline bool chkmax(T &x, const T2 &y) { return x < y ? x = y, 1 : 0; }
template<class T, class T2> inline bool chkmin(T &x, const T2 &y) { return x > y ? x = y, 1 : 0; }
const ll mod = 1e15 + 7;
#ifndef LOCAL
#define cerr if(false) cerr
#endif

typedef vector<vector<ll> > mat;

const ll MAX_N = 1e5 + 10;
ll n, m, k, q;

mat mult(const mat &a, const mat &b) {
    mat ret = mat(k, vector<ll>(k, mod));
    for(ll i = 0; i < k; i ++) {
        for(ll j = 0; j < k; j ++) {
            for(ll mid = 0; mid < k; mid ++) {
                chkmin(ret[i][j], a[i][mid] + b[mid][j]);
            }
        }
    }
    return ret;
}

mat prec[MAX_N];

vector<pair<ll, ll> > g[MAX_N];

signed main() {
    // ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
    cin >> k >> n >> m >> q;
    for(ll i = 0; i < m; i ++) {
        ll a, b, c;
        cin >> a >> b >> c;
        g[a].push_back({b, c});
    }
    for(ll i = 0; i < n; i += k) {
        prec[i / k] = mat(k, vector<ll>(k, mod));
        for(ll j = i; j < i + k; j ++) {
            for(auto it : g[j]) {
                chkmin(prec[i / k][j % k][it.first % k], it.second);
            }
        }
    }
    while(q --) {
        ll a, b;
        cin >> a >> b;
        if(a / k >= b / k) {
            cout << -1 << endl; 
            continue;
        } 
        mat ret = prec[a / k];
        a += k;
        while(a / k < b / k) {
            ret = mult(ret, prec[a / k]);
            a += k;
        }
        if(ret[a % k][b % k] == mod) {
            cout << -1 << endl; 
        } else {
            cout << ret[a % k][b % k] << endl;
        }
    }
    return 0;
}
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