제출 #535465

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
535465		fadi57	Fortune Telling 2 (JOI14_fortune_telling2)	C++14	100 / 100	694 ms	51064 KiB

fortune_telling2

#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
const int mx= 700 * 1000 + 3;
const int mx2=1e6+9;
typedef long long ll;
const int mod=998244353 ;
const int SQ=400;
const ll inf=1e9+10;
vector<int>cord,q[mx];
int seg[mx<<2],a[mx],b[mx],T[mx],aa[mx], bb[mx];int qq[mx];
tree<int,null_type,greater_equal<int>,rb_tree_tag,tree_order_statistics_node_update> s;
int sz;int t;int n,k;


void update(int node,int l,int r,int idx,int val){
   if(l==r){
    seg[node]=val;
    return;
   }
    int mid=(l+r)/2;
    if(idx<=mid){
    update(node*2,l,mid,idx,val);
   }else{
    update(node*2+1,mid+1,r,idx,val);
   }
   seg[node]=max(seg[node*2],seg[node*2+1]);
    return;
  }
  int query(int node,int l,int r,int left,int right){
         if(r<left||l>right){

            return -1;
      }
  if(l>=left&&r<=right){
        return seg[node];
     }
     int mid=(l+r)/2;
     return max(query(node*2,l,mid,left,right), query(node*2+1,mid+1,r,left,right));
     }


int fin(int x){
  return upper_bound(cord.begin(),cord.end(),x)-cord.begin();
  }

 int main(){
  cin>>n>>k;
 for(int i=0;i<n;i++){
    cin>>a[i]>>b[i];
    aa[i]=a[i];
    bb[i]=b[i];
    cord.push_back(a[i]);
    cord.push_back(b[i]);
  }
for(int i=0;i<k;i++){
     cin>>T[i];qq[i]=T[i];
     cord.push_back(T[i]);
  }
cord.erase(unique(cord.begin() , cord.end()) , cord.end()) ;sort(cord.begin(),cord.end());
sz=2*n+k;

for(int i=0;i<n;i++){
    a[i]=fin(a[i]);
    b[i]=fin(b[i]);
  }
   memset(seg,-1,sizeof(seg));
   for(int i=0;i<k;i++){

    T[i]=fin(T[i]);
    update(1,0,sz,T[i],i);
    }
  for(int i=0;i<n;i++){
  int idx=query(1,0,sz,min(a[i],b[i]),max(a[i],b[i])-1);
  q[idx+1].push_back(i);
  }

   ll ans=0;
   memset(seg,0,sizeof(seg));
   for (int i=k;i>=0;i--)
	{
	    if(i!=k){ s.insert(qq[i]);}
        for (int j:q[i])
		{   int parity;
			parity=s.order_of_key(max(aa[j],bb[j])-1);
		if(i==0)
    		parity ^= aa[j] < bb[j];
         if(parity%2){
            ans+=min(bb[j],aa[j]);

         }else{
          ans+=max(bb[j],aa[j]);

          }
    	//cout<<i<<" "<<j<<" "<<parity<<"\n";
    	}
    }
    	cout<<ans;
    	}

• Subtask 1: 4 / 4
• Subtask 2: 31 / 31
• Subtask 3: 65 / 65