제출 #957589

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
957589		vjudge1	로봇 (IOI13_robots)	C++17	0 / 100	1 ms	4444 KiB

robots

#include <bits/stdc++.h>
using namespace std;
#define pb push_back
#define se second
#define fi first
#include "robots.h"

int putaway(int a, int b, int t, int x[], int y[], int w[], int s[]) {
    vector <pair<int,int>> vec;
    for(int i=0;i<a;i++){
        vec.pb({x[i],1});
    }
    for(int i=0;i<t;i++){
        vec.pb({w[i],s[i]});
    }
    sort(vec.begin(),vec.end());
    sort(y,y+b,greater<int>());
    int l=0;
    int r=t+1;
    int ans=-1;
    while(l<r){
        int md=(l+r)>>1;
        bool ok=1;
        priority_queue<pair<int,int>>pq;
        for(int i=0;i<(int)vec.size();i++){
            if(vec[i].se==1){
                for(int j=0;j<md;j++){
                    if(pq.size())pq.pop();
                }
            }
            else{
                pq.push(vec[i]);
            }
        }
        priority_queue<int> pq2;
        while(pq.size()){
            pq2.push(pq.top().se);
            pq.pop();
        }
        for(int i=0;i<b;i++){
            if(!pq2.size())break;
            if(pq2.top()>=y[i]){
                ok=0;
                break;
            }
            else{
                for(int j=0;j<md;j++){
                    if(pq2.size())pq2.pop();
                }
            }
        }
        if(ok){
            ans=md;
            r=md;
        }
        else{
            l=md+1;
        }
    }
    return ans;
}

/*
---------------------------------------------------go get gold---------------------------------------------------------------------
- If u see the problem dp there's many option to iterate, u can iterate from 1- n or 1 - (possible max number)
- If u see the problem that has unexpected constraint u can divide it into 2 problem
- If u see the problem that can use binary search, then u can use binser + check condition
- in interactive problem there's many trick using binser
- kalo misal mau cari yang sama pake pernah ke visit ga bilangan itu
- kalo problem yang high itu biasanya optimisasi nya pake 2 array
- dp bisa aja kek kamu tenzing balls dimana optimisasi 2 dp
- janlup fibonacci
- kalo binser mending r = 3 * 1e18 aja
- kalo mau dibalik itu pake value nya tinggal diubah ke size
 */

• Subtask 1: 0 / 14
• Subtask 2: 0 / 14
• Subtask 3: 0 / 25
• Subtask 4: 0 / 37
• Subtask 5: 0 / 10