Submission #144405

#	Time	Username	Problem	Language	Result	Execution time	Memory
144405		qkxwsm	Vision Program (IOI19_vision)	C++14	100 / 100	28 ms	3068 KiB

vision

#include "vision.h"
#include <bits/stdc++.h>

using namespace std;

template<class T, class U>
void ckmin(T &a, U b)
{
	if (a > b) a = b;
}

template<class T, class U>
void ckmax(T &a, U b)
{
	if (a < b) a = b;
}

#define MP make_pair
#define PB push_back
#define LB lower_bound
#define UB upper_bound
#define fi first
#define se second
#define FOR(i, a, b) for (auto i = (a); i < (b); i++)
#define FORD(i, a, b) for (auto i = (a) - 1; i >= (b); i--)
#define SZ(x) ((int) (x).size())
#define ALL(x) (x).begin(), (x).end()

typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<pii> vpi;
typedef vector<pll> vpl;

int N, M, K, C, S[15];
vi tmp;

void construct_network(int n, int m, int k)
{
	//you want to see if the two squares with 1 are at distance K
	N = n; M = m; K = k;
	C = N * M;
	S[0] = C;
	FOR(i, 0, N + M - 1)
	{
		tmp.clear();
		FOR(x, 0, N)
		{
			int y = i - x;
			if (y < 0 || y >= M) continue;
			tmp.PB(x * M + y);
		}
		add_or(tmp);
		C++;
	}
	S[1] = C;
	FOR(i, -(M - 1), N)
	{
		tmp.clear();
		FOR(x, 0, N)
		{
			int y = x - i;
			if (y < 0 || y >= M) continue;
			tmp.PB(x * M + y);
		}
		add_or(tmp);
		C++;
	}
	S[2] = C;
	FOR(i, K, N + M - 1)
	{
		tmp.clear();
		tmp.PB(i + S[0]);
		tmp.PB((i - K) + S[0]);
		add_and(tmp);
		C++;
	}
	S[3] = C;
	FOR(i, -(M - 1) + K, N)
	{
		tmp.clear();
		tmp.PB(i + (M - 1) + S[1]);
		tmp.PB(i + (M - 1 - K) + S[1]);
		add_and(tmp);
		C++;
	}
	S[4] = C;
	tmp.clear();
	FOR(i, S[2], S[4])
	{
		tmp.PB(i);
	}
	add_or(tmp);
	C++;
	S[5] = C;
	//we need to make sure there's NOTHING strictly greater than K
	//precompute prefixes for x+ys
	tmp.clear();
	FOR(i, 0, N + M - 1)
	{
		tmp.PB(i + S[0]);
		add_or(tmp);
		C++;
	}
	S[6] = C;
	//precompute prefixes for x-ys
	tmp.clear();
	FOR(i, (-(M - 1)), N)
	{
		tmp.PB((i + M - 1) + S[1]);
		add_or(tmp);
		C++;
	}
	S[7] = C;
	FOR(i, K + 1, N + M - 1)
	{
		tmp.clear();
		tmp.PB(S[0] + i);
		tmp.PB(S[5] + (i - (K + 1)));
		add_and(tmp);
		C++;
	}
	S[8] = C;
	FOR(i, (-(M - 1) + K + 1), N)
	{
		tmp.clear();
		tmp.PB(S[1] + (M - 1 + i));
		tmp.PB(S[6] + (M - 1 + i - (K + 1)));
		add_and(tmp);
		C++;
	}
	S[9] = C;
	tmp.clear();
	FOR(i, S[7], S[9])
	{
		tmp.PB(i);
	}
	if (!tmp.empty())
	{
		add_or(tmp);
		C++;
		S[10] = C;
		//ok we nee S[4] to be good and S[9] to be bad.
		add_not(S[9]);
		tmp.clear();
		tmp.PB(S[4]);
		tmp.PB(S[10]);
		add_and(tmp);
	}
	else
	{
		tmp.PB(S[4]);
		add_or(tmp);
	}
	//try to see if there's a violation
	return;
}

• Subtask 1: 10 / 10
• Subtask 2: 11 / 11
• Subtask 3: 11 / 11
• Subtask 4: 15 / 15
• Subtask 5: 12 / 12
• Subtask 6: 8 / 8
• Subtask 7: 14 / 14
• Subtask 8: 19 / 19