제출 #540529

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
540529		PedroBigMan	자리 배치 (IOI18_seats)	C++14	0 / 100	4043 ms	241348 KiB

seats

/*
Author of all code: Pedro BIGMAN Dias
Last edit: 15/02/2021
*/
#pragma GCC optimization ("O3")
#pragma GCC optimization ("unroll-loops")
#pragma GCC optimize("Ofast")
#include <iostream>
#include <vector>
#include <cmath>
#include <algorithm>
#include <string>
#include <map>
#include <unordered_map>
#include <set>
#include <unordered_set>
#include <queue>
#include <deque>
#include <list>
#include <iomanip>
#include <stdlib.h>
#include <time.h>
#include <cstring>
#include "seats.h"
using namespace std;
typedef int ll;
typedef unsigned long long int ull;
typedef long double ld;
#define REP(i,a,b) for(ll i=(ll) a; i<(ll) b; i++)
#define pb push_back
#define mp make_pair
#define pl pair<ll,ll>
#define ff first
#define ss second
#define whole(x) x.begin(),x.end()
#define DEBUG(i) cout<<"Pedro Is The Master "<<i<<endl
#define INF 500000000LL
#define EPS 0.00000001
#define pi 3.14159
#define VV(vvvv,NNNN,xxxx); REP(i,0,NNNN) {vvvv.pb(xxxx);}
ll mod=1000000007LL;

template<class A=ll> 
void Out(vector<A> a) {REP(i,0,a.size()) {cout<<a[i]<<" ";} cout<<endl;}

template<class A=ll>
void In(vector<A> &a, ll N) {A cur; REP(i,0,N) {cin>>cur; a.pb(cur);}} 

class ST_max
{
    public:
    ll N;
    
    class SV //seg value
    {
        public:
        ll a; 
        SV() {a=0LL;}
        SV(ll x) {a=x;}
        
        SV operator & (SV X) {SV ANS(max(a,X.a)); return ANS;}
    };
    
    SV neuts; 
	
	      
    SV upval(ll c) //how lazy values modify a seg value inside a node, c=current node
    {
        return p[c];
    }
    
    vector<SV> p;
    vector<pl> range;
    
    ST_max() {N=0LL;}
    ST_max(vector<ll> arr)
    {
        N = (ll) 1<<(ll) ceil(log2(arr.size()));
        REP(i,0,2*N) {range.pb(mp(0LL,0LL));}
        REP(i,0,N) {p.pb(neuts);}
        REP(i,0,arr.size()) {SV X(arr[i]); p.pb(X); range[i+N]=mp(i,i);}
        REP(i,arr.size(),N) {p.pb(neuts); range[i+N]=mp(i,i);}
        ll cur = N-1;
        while(cur>0)
        {
            p[cur]=p[2*cur]&p[2*cur+1];
            range[cur]=mp(range[2*cur].ff,range[2*cur+1].ss);
            cur--;
        }
	}
    
    SV query(ll a,ll b, ll c=1LL) //range [a,b], current node. initially: query(a,b)
    {
        ll x=range[c].ff; ll y=range[c].ss;
        if(y<a || x>b) {return neuts;}
        if(x>=a && y<=b) {return upval(c);}
        SV ans = query(a,b,2*c)&query(a,b,2*c+1);
        return ans;
    }
    
    void update(SV s, ll a, ll b, ll c=1LL) //update LV, range [a,b], current node, current range. initially: update(s,a,b)
    {
        ll x=range[c].ff; ll y=range[c].ss;
        if(y<a || x>b) {return ;}
        if(x>=a && y<=b) 
        {
            p[c]=s;
            return;
        }
        update(s,a,b,2*c); update(s,a,b,2*c+1);
        p[c]=upval(2*c)&upval(2*c+1);
    }
};

class ST_min
{
    public:
    ll N;
    
    class SV //seg value
    {
        public:
        ll a; 
        SV() {a=INF;}
        SV(ll x) {a=x;}
        
        SV operator & (SV X) {SV ANS(min(a,X.a)); return ANS;}
    };
    
    SV neuts; 
	
    SV upval(ll c) //how lazy values modify a seg value inside a node, c=current node
    {
        return p[c];
    }
    
    vector<SV> p;
    vector<pl> range;
    
    ST_min() {N=0LL;}
    ST_min(vector<ll> arr)
    {
        N = (ll) 1<<(ll) ceil(log2(arr.size()));
        REP(i,0,2*N) {range.pb(mp(0LL,0LL));}
        REP(i,0,N) {p.pb(neuts);}
        REP(i,0,arr.size()) {SV X(arr[i]); p.pb(X); range[i+N]=mp(i,i);}
        REP(i,arr.size(),N) {p.pb(neuts); range[i+N]=mp(i,i);}
        ll cur = N-1;
        while(cur>0)
        {
            p[cur]=p[2*cur]&p[2*cur+1];
            range[cur]=mp(range[2*cur].ff,range[2*cur+1].ss);
            cur--;
        }
	}
    
    SV query(ll a,ll b, ll c=1LL) //range [a,b], current node. initially: query(a,b)
    {
        ll x=range[c].ff; ll y=range[c].ss;
        if(y<a || x>b) {return neuts;}
        if(x>=a && y<=b) {return upval(c);}
        SV ans = query(a,b,2*c)&query(a,b,2*c+1);
        return ans;
    }
    
   	void update(SV s, ll a, ll b, ll c=1LL) //update LV, range [a,b], current node, current range. initially: update(s,a,b)
    {
        ll x=range[c].ff; ll y=range[c].ss;
        if(y<a || x>b) {return ;}
        if(x>=a && y<=b) 
        {
            p[c]=s;
            return;
        }
        update(s,a,b,2*c); update(s,a,b,2*c+1);
        p[c]=upval(2*c)&upval(2*c+1);
    }
};

ll h,w,N;
ST_max H_max,W_max; ST_min H_min,W_min;
vector<ll> ans; ll t_ans;

void give_initial_chart(int hh, int ww, vector<int> RR, vector<int> CC) 
{
 	h=hh; w=ww; N=h*w; ST_max HH_max(RR); H_max=HH_max; ST_min HH_min(RR); H_min=HH_min; ST_max WW_max(CC); W_max=WW_max; ST_min WW_min(CC); W_min=WW_min;
	t_ans=0LL; ll max_l,min_l,max_c,min_c;
	REP(i,0,N)
	{
		max_l=H_max.query(0,i).a;
		min_l=H_min.query(0,i).a;
		max_c=W_max.query(0,i).a;
		min_c=W_min.query(0,i).a;
		if((max_c-min_c)*(max_l-min_l)==i+1) {ans.pb(1); t_ans++;} else {ans.pb(0);}
	}
}

int swap_seats(int a, int b) 
{
  	ll line_a = H_max.query(a,a).a; ll col_a = W_max.query(a,a).a;
  	ll line_b = H_max.query(b,b).a; ll col_b = W_max.query(b,b).a;
	H_min.update(line_b,a,a); H_max.update(line_b,a,a);
	H_min.update(line_a,b,b); H_max.update(line_a,b,b);
	W_min.update(col_b,a,a); W_max.update(col_b,a,a);
	W_min.update(col_a,b,b); W_max.update(col_a,b,b);
	ll max_l,min_l,max_c,min_c; ll oldval,newval;
	REP(i,a,b+1)
	{
		max_l=H_max.query(0,i).a;
		min_l=H_min.query(0,i).a;
		max_c=W_max.query(0,i).a;
		min_c=W_min.query(0,i).a;
		oldval=ans[i];
		if((max_c-min_c)*(max_l-min_l)==i+1) {newval=1;} else {newval=0;}
		ans[i]=newval;
		t_ans+=(newval-oldval);
	}
	return t_ans;
}

컴파일 시 표준 에러 (stderr) 메시지

seats.cpp:5: warning: ignoring '#pragma GCC optimization' [-Wunknown-pragmas]
    5 | #pragma GCC optimization ("O3")
      | 
seats.cpp:6: warning: ignoring '#pragma GCC optimization' [-Wunknown-pragmas]
    6 | #pragma GCC optimization ("unroll-loops")
      |

• Subtask 1: 0 / 5
• Subtask 2: 0 / 6
• Subtask 3: 0 / 20
• Subtask 4: 0 / 6
• Subtask 5: 0 / 33
• Subtask 6: 0 / 30