이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
// Code by Parsa Eslami
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define ll long long
#define pii pair<int,int>
#define pll pair<ll,ll>
#define FOR(i,a,b) for(int i=a;i<=b;i++)
#define FORR(i,a,b) for(int i=a;i>=b;i--)
#define F first
#define S second
#define maxm(a,b) a=max(a,b)
#define minm(a,b) a=min(a,b)
#define bit(i,j) ((j>>i)&1)
#define pb push_back
#define all(x) x.begin(),x.end()
#define SZ(x) (int)x.size()
#define err(x) cerr<<#x<<" : "<<x<<'\n'
#define dmid int mid=(r+l)/2
#define lc 2*id
#define rc 2*id+1
using namespace std;
using namespace __gnu_pbds;
template<typename T>
using ordered_set=tree<T,null_type,less<T>,rb_tree_tag,tree_order_statistics_node_update>;
const int N=2e5+4;
const int inf=1e9+1;
ordered_set<pii> seg[4*N];
void upd(int id,int l,int r,int ind,pii x){
if(ind>=r||ind<l) return;
seg[id].insert(x);
if(l+1==r) return;
dmid;
upd(lc,l,mid,ind,x);
upd(rc,mid,r,ind,x);
}
int get(int id,int l,int r,int st,int en,int y){
if(st>=r||en<=l) return 0;
if(st<=l&&en>=r) return (seg[id].order_of_key({inf,0})-seg[id].order_of_key({y,0}));
dmid;
return (get(lc,l,mid,st,en,y)+get(rc,mid,r,st,en,y));
}
int main(){
iostream::sync_with_stdio(0); cin.tie(0);
int n,q; cin>>n>>q;
int a[n+1],b[n+1],c[n+1];
FOR(i,1,n){ cin>>a[i]>>b[i]; c[i]=a[i]+b[i];}
int X[q+1],Y[q+1],S[q+1];
FOR(i,1,q) cin>>X[i]>>Y[i]>>S[i];
vector<pii> vc;
FOR(i,1,n) vc.pb({a[i],2*i});
FOR(i,1,q) vc.pb({X[i],2*i+1});
sort(all(vc));
int comp=0;
FOR(i,0,SZ(vc)-1){
if(i==0||vc[i].F!=vc[i-1].F) comp++;
if(vc[i].S&1) X[vc[i].S/2]=comp;
else a[vc[i].S/2]=comp;
}
int ans[q+1]={};
vector<pii> vc0;
vector<pii> vc1;
FOR(i,1,n) vc0.pb({c[i],i});
FOR(i,1,q) vc1.pb({S[i],i});
sort(all(vc0)); sort(all(vc1));
while(!vc1.empty()){
while(!vc0.empty()&&vc0.back().F>=vc1.back().F){
int i=vc0.back().S;
upd(1,1,n+q+1,a[i],{b[i],i});
vc0.pop_back();
}
int i=vc1.back().S;
ans[i]=get(1,1,n+q+1,X[i],n+q+1,Y[i]);
vc1.pop_back();
}
FOR(i,1,q) cout<<ans[i]<<'\n';
return 0;
}
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