Submission #62272

#	Time	Username	Problem	Language	Result	Execution time	Memory
62272		Benq	Cambridge (info1cup18_cambridge)	C++11	100 / 100	627 ms	33168 KiB

cambridge

#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 = 1<<17;

template<class T, int SZ> struct LazySegTree {
    T sum[2*SZ], mn[2*SZ], lazy[2*SZ]; // set SZ to a power of 2
    
    LazySegTree() {
        memset (sum,0,sizeof sum);
        memset (mn,0,sizeof mn);
        memset (lazy,0,sizeof lazy);
    }
    
    void push(int ind, int L, int R) {
        sum[ind] += (R-L+1)*lazy[ind];
        mn[ind] += lazy[ind];
        if (L != R) lazy[2*ind] += lazy[ind], lazy[2*ind+1] += lazy[ind];
        lazy[ind] = 0;
    }
    
    void pull(int ind) {
        sum[ind] = sum[2*ind]+sum[2*ind+1];
        mn[ind] = min(mn[2*ind],mn[2*ind+1]);
    }
    
    void build() {
        F0Rd(i,SZ) pull(i);
    }
    
    T qsum(int lo, int hi, int ind = 1, int L = 0, int R = SZ-1) {
        push(ind,L,R);
        if (lo > R || L > hi) return 0;
        if (lo <= L && R <= hi) return sum[ind];
        
        int M = (L+R)/2;
        return qsum(lo,hi,2*ind,L,M) + qsum(lo,hi,2*ind+1,M+1,R);
    }

    T qmin(int lo, int hi, int ind = 1, int L = 0, int R = SZ-1) {
        push(ind,L,R);
        if (lo > R || L > hi) return INF;
        if (lo <= L && R <= hi) return mn[ind];
        
        int M = (L+R)/2;
        return min(qmin(lo,hi,2*ind,L,M), qmin(lo,hi,2*ind+1,M+1,R));
    }
    
    void upd(int lo, int hi, ll inc, int ind = 1, int L = 0, int R = SZ-1) {
        push(ind,L,R);
        if (hi < L || R < lo) return;
        if (lo <= L && R <= hi) {
            lazy[ind] = inc;
            push(ind,L,R);
            return;
        }
        
        int M = (L+R)/2;
        upd(lo,hi,inc,2*ind,L,M); upd(lo,hi,inc,2*ind+1,M+1,R);
        pull(ind);
    }
};

LazySegTree<ll,MX> L;

int N,M,key[MX], lst[MX],T[MX],D[MX],ans[MX];
map<pi,int> m;

int main() {
    ios_base::sync_with_stdio(0); cin.tie(0);
    cin >> N >> M;
    FOR(i,1,N+1) {
        cin >> T[i] >> D[i];
        m[{D[i],i}] = 0;
    }
    int co = 0; for (auto& a: m) a.s = co++;
    FOR(i,1,N+1) key[i] = m[{D[i],i}];
    int r = 0;
    L.upd(0,MX-1,MOD);
    FOR(i,1,N+1) {
        while (L.qmin(0,MX-1) > 0 && r <= N) {
            r ++;
            if (r != N+1) {
                L.upd(key[r],key[r],D[r]-MOD);
                L.upd(key[r],MX-1,-T[r]);
            }
        }
        L.upd(key[i],key[i],MOD-D[i]);
        L.upd(key[i],MX-1,T[i]);
        ans[i] = r-1;
    }
    F0R(i,M) {
        int x,y; cin >> x >> y;
        cout << (ans[x] >= y) << "\n";
    }
}

/* 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: 15 / 15
• Subtask 2: 25 / 25
• Subtask 3: 15 / 15
• Subtask 4: 45 / 45