Submission #673526

#	Time	Username	Problem	Language	Result	Execution time	Memory
673526		Carmel_Ab1	Examination (JOI19_examination)	C++17	100 / 100	1388 ms	550756 KiB

examination

/*
#pragma GCC target ("avx2")
#pragma GCC optimization ("O3")
#pragma GCC optimization ("unroll-loops")
 */
#include<bits/stdc++.h>
//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>

//using namespace __gnu_pbds;
using namespace std;

typedef long double ld;
typedef long long ll;
typedef unsigned long long ull;
typedef vector<int>vi;
typedef vector<vector<int>>vvi;
typedef vector<ll>vl;
typedef vector<vl> vvl;
typedef pair<int,int>pi;
typedef pair<ll,ll> pl;
typedef vector<pl> vpl;
typedef vector<ld> vld;
typedef pair<ld,ld> pld;
typedef vector<pi> vpi;

//typedef tree<ll, null_type, less_equal<ll>,rb_tree_tag,tree_order_statistics_node_update> ordered_set;
template<typename T> ostream& operator<<(ostream& os, vector<T>& a){os<<"[";for(int i=0; i<ll(a.size()); i++){os << a[i] << ((i!=ll(a.size()-1)?" ":""));}os << "]\n"; return os;}

#define all(x) x.begin(),x.end()
#define YES out("YES")
#define NO out("NO")
#define out(x){cout << x << "\n"; return;}
#define outfl(x){cout << x << endl;return;}
#define GLHF ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL)
#define print(x){for(auto ait:x) cout << ait << " "; cout << "\n";}
#define pb push_back
#define umap unordered_map


template<typename T>
void read(vector<T>& v){
    int n=v.size();
    for(int i=0; i<n; i++)
        cin >> v[i];
}
template<typename T>
vector<T>UNQ(vector<T>a){
    vector<T>ans;
    for(T t:a)
        if(ans.empty() || t!=ans.back())
            ans.push_back(t);
    return ans;
}



void solve();
int main(){
    GLHF;
    int t=1;
    //cin >> t;
    while(t--)
        solve();
}
struct seg{
    seg* lp=0,*rp=0;
    int l,r,m;
    int sum=0;

    seg(int l,int r):l(l),r(r),m((l+r)/2){}

    void expand(){
        if(!lp)lp=new seg(l,m);
        if(!rp)rp=new seg(m,r);
    }
    void upd(int i,int v){
        sum+=v;
        if(l+1==r)
            return;
        expand();
        if(i<m)
            lp->upd(i,v);
        else
            rp->upd(i,v);
    }

    int qur(int a,int b){
        if(b<=l || r<=a)
            return 0;
        if(a<=l && r<=b)
            return sum;
        expand();
        return lp->qur(a,b) + rp->qur(a,b);
    }
};

/*
 Basically what we do is split the queries into 2 types:
 type (1): queries where A+B>=C: then we only need to count S>=A, T>=B, (and then it applies that S+T>=A+B>=C).
 So we only need to count how many fulfil S>=A, T>=B.
 This can be done offline - sorting queries and pairs by S/A, then querying a segment tree on [B,inf].

 type (2): queries where A+B<C:
 Now try to visualize pairs of (S[i],T[i]) on the cartesian plane.
 The restrictions of the queries can be modeled in the following way:

  \   |
   \  |
    \ |"This area is what we want"
     \|
      |\
----------------------------------------------- "limit A"
      |   \
      |    \"limit C"
     "limit B"

 Then we can count how many (S+T>=C AND T>=B) - (S+T>=C AND A<S).

 This can be done similarly to type (1) queries.

 Also see: https://codeforces.com/blog/entry/66022?#comment-501375
 */
void solve() {
    int n,q;
    cin >> n >> q;
    vvl a(n,vl(2));
    for(int i=0 ;i<n; i++)
        cin >> a[i][0] >> a[i][1];

    vvl queries(q,vl(4));
    for(int i=0; i<q; i++){
        cin >>queries[i][0] >> queries[i][1] >> queries[i][2];
        queries[i][3]=i;
    }
    sort(all(a));
    sort(all(queries));

    vi ans(q);
    seg ST(0,1e9 +10);

    for(int i=0; i<n; i++)
        ST.upd(a[i][1],1);

    for(int i=0,j=0; i<q; i++){
        while(j<n && a[j][0]<queries[i][0])
            ST.upd(a[j++][1],-1);
        if(queries[i][0]+queries[i][1]>=queries[i][2])
            ans[queries[i][3]] = ST.qur(queries[i][1],1e9 + 2);
    }

    sort(all(a),[&](vl v1,vl v2){
        return v1[0]+v1[1] < v2[0]+v2[1];
    });
    sort(all(queries),[&](vl v1,vl v2){
       return v1[2]<v2[2];
    });
    ST = seg(0,1e9 + 10);
    for(int i=0; i<n;i++)
        ST.upd(a[i][1],1);

    for(int i=0,j=0; i<q; i++){
        ll A=queries[i][0],B=queries[i][1],C=queries[i][2],ix=queries[i][3];
        if(A+B>=C)
            continue;
        while(j<n && a[j][0]+a[j][1]<C)
            ST.upd(a[j][1],-1), j++;
        ans[ix] = ST.qur(B,1e9 + 2);
    }


    ST = seg(0,1e9 + 10);
    for(int i=0; i<n;i++)
        ST.upd(a[i][0],1);
    for(int i=0,j=0; i<q; i++){
        ll A=queries[i][0],B=queries[i][1],C=queries[i][2],ix=queries[i][3];
        if(A+B>=C)
            continue;
        while(j<n && a[j][0]+a[j][1]<C)
            ST.upd(a[j][0],-1), j++;
        ans[ix] -= ST.qur(0,A);
    }


    for(int x:ans)
        cout << x << "\n";
}

• Subtask 1: 2 / 2
• Subtask 2: 20 / 20
• Subtask 3: 21 / 21
• Subtask 4: 57 / 57