제출 #527493

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
527493		zaneyu	On the Grid (FXCUP4_grid)	C++17	100 / 100	14 ms	336 KiB

grid

#include "grid.h"
#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;
 
//#pragma GCC optimize("unroll-loops,no-stack-protector")
//order_of_key #of elements less than x
// find_by_order kth element
using ll = long long;
using ld = long double;
using pii = pair<ll,ll>;
typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> indexed_set;
#define f first
#define s second
#define pb push_back
#define REP(i,n) for(int i=0;i<n;i++)
#define REP1(i,n) for(int i=1;i<=n;i++)
#define FILL(n,x) memset(n,x,sizeof(n))
#define ALL(_a) _a.begin(),_a.end()
#define sz(x) (int)x.size()
#define SORT_UNIQUE(c) (sort(c.begin(),c.end()), c.resize(distance(c.begin(),unique(c.begin(),c.end()))))
#define MP make_pair
const ll INF64=4e17;
const int INF=0x3f3f3f3f;
const ll MOD=1e6+7;
const ld PI=acos(-1);
const ld eps=1e-9;
#define lowb(x) x&(-x)
#define MNTO(x,y) x=min(x,(__typeof__(x))y)
#define MXTO(x,y) x=max(x,(__typeof__(x))y)
template<typename T1,typename T2>
ostream& operator<<(ostream& out,pair<T1,T2> P){
    out<<P.f<<' '<<P.s;
    return out;
}
template<typename T>
ostream& operator<<(ostream& out,vector<T> V){
    REP(i,sz(V)) out<<V[i]<<((i!=sz(V)-1)?" ":"");
    return out;
}
ll mult(ll a,ll b){
    return ((a%MOD)*(b%MOD))%MOD;
}
ll mypow(ll a,ll b){
    if(b<=0) return 1;
    ll res=1LL;
    while(b){
        if(b&1) res=mult(res,a);
        a=mult(a,a);
        b>>=1;
    }
    return res;
}
const int maxn=1e3+5;
int ok[maxn];
std::vector<int> SortDisks(int n) {
	vector<int> v,ans(n);
	REP(i,n) v.pb(i),ok[i]=-1;
	int cur=PutDisks(v);
	while(sz(v)){
		vector<int> qq;
		int x=v.back(),p=0;
		for(int i=sz(v)-1;i>=1;i--) v[i]=v[i-1];
		v[0]=x;
		REP(i,n){
			if(ok[i]!=-1){
				qq.pb(ok[i]);
			}
			else qq.pb(v[p++]);
		}
		int b=PutDisks(qq);
		if(b>=cur){
			int t=v[0];
			for(int i=n-1;qq[i]!=v[0];i--) b--;
			ans[t]=b,ok[b-1]=t;
			REP(i,sz(v)-1) v[i]=v[i+1]; 
			v.pop_back();
		}
		else cur=b;
	}
	return ans;
}

• Subtask 1: 12 / 12
• Subtask 2: 31 / 31
• Subtask 3: 57 / 57