# | Time | Username | Problem | Language | Result | Execution time | Memory |
---|---|---|---|---|---|---|---|
557831 | hibiki | Examination (JOI19_examination) | C++11 | 342 ms | 22928 KiB |
This submission is migrated from previous version of oj.uz, which used different machine for grading. This submission may have different result if resubmitted.
#include<bits/stdc++.h>
using namespace std;
#define PB push_back
#define F first
#define S second
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
template <typename T> using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
int n,q;
int sA[100005],sB[100005];
int qA[100005],qB[100005],qC[100005],ans[100005];
vector<pair<int,pair<int,int> > > poi;
int main()
{
scanf("%d %d",&n,&q);
for(int i=0;i<n;i++)
{
scanf("%d %d",&sA[i],&sB[i]);
poi.PB({ -(sA[i]+sB[i]) , {0,i} });
}
for(int i=0;i<q;i++)
{
scanf("%d %d %d",&qA[i],&qB[i],&qC[i]);
poi.PB({ -max(qA[i]+qB[i],qC[i]) , {1,i} });
}
sort(poi.begin(),poi.end());
int all=0;
ordered_set<pair<int,int> > oA,oB;
for(int i=0;i<poi.size();i++)
{
int ty=poi[i].S.F;
int idx=poi[i].S.S;
if(!ty)
{
all++;
oA.insert({sA[idx],idx});
oB.insert({sB[idx],idx});
}
else
{
ans[idx] = all - oA.order_of_key({qA[idx],0}) - oB.order_of_key({qB[idx],0});
}
}
for(int i=0;i<q;i++)
printf("%d\n",ans[i]);
return 0;
}
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