Submission #1135490

#	Time	Username	Problem	Language	Result	Execution time	Memory
1135490		vibeduck	Vision Program (IOI19_vision)	C++20	0 / 100	11 ms	1092 KiB

vision

#include "vision.h"
#include <bits/stdc++.h>
using namespace std;

typedef long double ld;
typedef long long ll;
typedef pair<ll,ll> pll;
typedef pair<int,int> pii;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef vector<bool> vb;
typedef vector<vector<int>> vvi;
typedef vector<vector<bool>> vvb;
typedef vector<vector<ll>> vvll;
typedef vector<string> vs;
typedef vector<vector<string>> vvs;
typedef vector<char> vc;
typedef vector<vector<char>> vvc;
typedef map<int, int> mii;
typedef unordered_map<int, int> umii;

/*
0 -> H - 1 : is one in this row
H -> H + W - 1 : is one in this col

H + W -> both in same row
H + W + 1 -> H + W + W - 1 : in adj columns for I
H + W + W : in adj columns
H + W + W + 1 : case 1 ans

H + W + W + 2 : both in same col
H + W + W + 3 -> H + H + W + W + 1 : in adj rows for I
H + H + W + W + 2 : in adj rows
H + H + W + W + 3 : case 2 ans

H + H + W + W + 4 : overall ans
*/

void construct_network(int H, int W, int K) {
	int cnt = 0;

	// + H
	for (int i = 0; i < H; i++) {
		vi cur;
		for (int j = 0; j < W; j++) cur.push_back(i * W + j);
		add_or(cur);
		cnt++;
	}
	// + W
	for (int i = 0; i < W; i++) {
		vi cur;
		for (int j = 0; j < H; j++) cur.push_back(j * W + i);
		add_or(cur);
		cnt++;
	}

	// both in same row
	// + 1
	vi x;
	for (int i = 0; i < H; i++) x.push_back(H * W + i);
	add_xor(x);
	cnt++;
	
	/*

	// both in adj col for I
	// + W - 1
	x.clear();
	for (int i = 0; i < W - 1; i++) {
		add_and({H * W + H + i, H * W + H + i + 1});
		cnt++;
	}

	// both actually in adj col
	// + 1
	vi t;
	for (int i = H + W + 1; i <= H + W + W - 1; i++) t.push_back(H * W + i);
	add_xor(t); 
	cnt++;

	// case 1 ans
	// + 1
	add_and({H + W, H + W + W});
	cnt++;



	

	// both in same col
	// + 1
	x.clear();
	for (int i = H; i <= H + W - 1; i++) x.push_back(H * W + i);
	add_xor(x);
	cnt++;

	// both in adj row for I
	// + H - 1
	x.clear();
	for (int i = 0; i < H - 1; i++) {
		add_and({H * W + i, H * W + i + 1});
		cnt++;
	}

	// both actually in adj row
	// + W - 1
	t.clear();
	for (int i = H + W + W + 3; i <= H + H + W + W + 1; i++) t.push_back(H * W + i);
	add_xor(t); 
	cnt++;

	// case 2 ans
	add_and({H + H + W + W + 2, H + W + W + 2});
	cnt++;

	// overall ans
	add_or({H + W + W + 1, H + H + W + W + 3});

	*/	
}

• Subtask 1: 0 / 10
• Subtask 2: 0 / 11
• Subtask 3: 0 / 11
• Subtask 4: 0 / 15
• Subtask 5: 0 / 12
• Subtask 6: 0 / 8
• Subtask 7: 0 / 14
• Subtask 8: 0 / 19