# | Time | Username | Problem | Language | Result | Execution time | Memory |
---|---|---|---|---|---|---|---|
847530 | Lib | Closing Time (IOI23_closing) | C++17 | 63 ms | 23492 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 "closing.h"
#include <bits/stdc++.h>
using namespace std;
long long pos[200003];
long long cost1[200003];
long long cost2[200003];
long long pfx1[200003];
long long pfx2[200003];
long long check[200003];
long long maxroll[200003]; //maxroll[i]=a <==> de "tra no" 1 khoang toi da =a, can rollback it nhat i vi tri tu loai 1 xuong loai 0
long long minupd[200003]; //minupt[i]=k <==> voi 1 budget<=i; upgrade duoc toi da k vi tri tu loai 1 len loai 2
struct cus{
bool operator()(const long long &x, const long long &y){
return x>y;
}
};
int max_score(int N, int X, int Y, long long K, std::vector<int> U, std::vector<int> V, std::vector<int> W)
{
int n=N;
int x=X;
int y=Y;
long long k=K;
for(int i=0;i<=200001;i++){
cost1[i]=0;
cost2[i]=0;
pfx1[i]=0;
pos[i]=0;
check[i]=0;
}
for(int i=0;i<n-1;i++){
pos[i+1]=pos[i]+W[i];
}
for(int i=0;i<n;i++){
cost1[i]=min(abs(pos[i]-pos[x]),abs(pos[i]-pos[y]));
cost2[i]=max(abs(pos[i]-pos[x]),abs(pos[i]-pos[y]))-cost1[i];
pfx1[i+1]=pfx1[i]+cost1[i];
}
priority_queue <pair <long long,int> > pq;
long long mcost=k;
long long ans=0,tans=0;
int c;
int ccnt=0;
int tpos;
//phase 1: thi nang cap cac diem len toi da lv1
for(int i=0;i<n;i++){
pq.push({-cost1[i],i});
}
ans=0;
while(mcost>0&&!pq.empty()){
mcost+=pq.top().first;
ans++;
pq.pop();
}
if(mcost<0){
ans--;
}
tans=max(tans,ans);
while(!pq.empty()){
pq.pop();
}
//phase 2: co ca loai 1 lan 2. Khi do, thang loai 2 chi duoc accept khi o day da co 1 roi
mcost=k;
mcost-=(pfx1[y+1]-pfx1[x]);
if(mcost>0){
ans=0;
for(int i=0;i<n;i++){
if(i<x||i>y){
pq.push({-cost1[i],i});
}else{
pq.push({0,i});
}
}
while(!pq.empty()&&mcost>0){
mcost+=pq.top().first;
if(mcost>=0){
ans++;
tpos=pq.top().second;
pq.pop();
check[tpos]++;
//cout<<tpos<<" "<<check[tpos]<<" "<<-pq.top().first<<"\n";
if(check[tpos]==1){
pq.push({-cost2[tpos],tpos});
}
}
}
tans=max(tans,ans);
}
while(!pq.empty()){
pq.pop();
}
return tans;
}
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