Submission #57084

#	Time	Username	Problem	Language	Result	Execution time	Memory
57084		Benq	Pinball (JOI14_pinball)	C++14	100 / 100	901 ms	55736 KiB

pinball

#include <bits/stdc++.h>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>

using namespace std;
using namespace __gnu_pbds;
 
typedef long long ll;
typedef long double ld;
typedef complex<ld> cd;

typedef pair<int, int> pi;
typedef pair<ll,ll> pl;
typedef pair<ld,ld> pd;

typedef vector<int> vi;
typedef vector<ld> vd;
typedef vector<ll> vl;
typedef vector<pi> vpi;
typedef vector<pl> vpl;
typedef vector<cd> vcd;

template <class T> using Tree = tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update>;

#define FOR(i, a, b) for (int i=a; i<(b); i++)
#define F0R(i, a) for (int i=0; i<(a); i++)
#define FORd(i,a,b) for (int i = (b)-1; i >= a; i--)
#define F0Rd(i,a) for (int i = (a)-1; i >= 0; i--)

#define sz(x) (int)(x).size()
#define mp make_pair
#define pb push_back
#define f first
#define s second
#define lb lower_bound
#define ub upper_bound
#define all(x) x.begin(), x.end()

const int MOD = 1000000007;
const ll INF = 1e18;
const int MX = 100001;

template<class T, int SZ> struct Seg {
    T seg[2*SZ], MN = INF;
    
    Seg() {
        F0R(i,2*SZ) seg[i] = INF;
    }
    
    T comb(T a, T b) { return min(a,b); } // easily change this to min or max
    
    void upd(int p, T value) {  // set value at position p
        for (seg[p += SZ] = value; p > 1; p >>= 1)
            seg[p>>1] = comb(seg[(p|1)^1],seg[p|1]); // non-commutative operations
    }
    
    void build() {
        F0Rd(i,SZ) seg[i] = comb(seg[2*i],seg[2*i+1]);
    }
    
    T query(int l, int r) {  // sum on interval [l, r]
        T res1 = MN, res2 = MN; r++;
        for (l += SZ, r += SZ; l < r; l >>= 1, r >>= 1) {
            if (l&1) res1 = comb(res1,seg[l++]);
            if (r&1) res2 = comb(seg[--r],res2);
        }
        return comb(res1,res2);
    }
};

Seg<ll,1<<19> L, R;

int N,M;
map<int,int> m;
vector<vi> v;

void input() {
    ios_base::sync_with_stdio(0); cin.tie(0);
    cin >> M >> N;
    m[1] = 0;
    m[N+1] = 0;
    F0R(i,M) {
        int A,B,C,D; cin >> A >> B >> C >> D;
        v.pb({A,B,C,D});
        m[A] = m[B+1] = m[C] = 0;
    }
    int co = 0;
    for (auto& a: m) a.s = co++;
    L.upd(0,0);
    R.upd(m[N+1]-1,0);
}

int main() {
    input();
    ll ans = INF;
    // cout << L.query(1,1) << "\n";
    for (auto a: v) {
        int l = m[a[0]], r = m[a[1]+1]-1;
        int mid = m[a[2]];
        // cout << l << " " << mid << " " << r << "\n";
        ans = min(ans,a[3]+L.query(l,r)+R.query(l,r));
        L.upd(mid,min(L.query(l,r)+a[3],L.query(mid,mid)));
        R.upd(mid,min(R.query(l,r)+a[3],R.query(mid,mid)));
    }
    if (ans == INF) cout << -1;
    else cout << ans;
}

/* Look for:
* the exact constraints (multiple sets are too slow for n=10^6 :( ) 
* special cases (n=1?)
* overflow (ll vs int?)
* array bounds
*/

• Subtask 1: 11 / 11
• Subtask 2: 18 / 18
• Subtask 3: 22 / 22
• Subtask 4: 49 / 49