/**
* Wow. Idk how to express my shock when i saw that interactive problems can be like
* this! So we basically have to write a program that outputs the correct 0/1 across
* all images. First thought: we can't do compare each pair of rows, this takes about
* 19900 (>10000) calls.
* We don't have "for" or "if" statements... Hmm. I don't wanna do int addition /
* subtraction using logic gates too... Well, in general we don't have much calls
* available, so perhaps in a lot of calls the num of input isn't small. Nah...
*
* I'd like to note down my final thought. Suppose that c_1, c_2, ..., c_m are 0/1
* bits representing whether column i has a black cell. Let's put away the case of
* only one 1. Let d_i := c_1 ^ c_2 ^ ... ^ c_i, so the number of 1s in (d_i) is
* the horizontal dist. Similarly we can do this for the vertical direction, then
* it suffices to count the num of 1s in some cells and check if it's K.
* Unfortunately, the only sol i have is to do a dp which requires an additional
* K(H+W) cells. Too much... (I'll submit a brute-force version in impl1)
*
* This think this is a problem where scoring partials is easy, but getting full is
* hard.
*
* Number of calls: 2HW
* Number of reads: idk, but too many
* Implementation 1 (Solves [S1-3], [S5], [S6])
*/
#include <bits/stdc++.h>
#include "vision.h"
typedef std::vector<int> vec;
void construct_network(int h, int w, int K) {
if (h * w <= 900) {
vec list;
for (int i = 0; i < h; i++) {
for (int j = 0; j < w; j++) {
vec ors;
for (int delta_i = -K; delta_i <= K; delta_i++) {
std::cerr << "[debug] " << i << ' ' << j << ' ' << delta_i << std::endl;
int delta_j = K - std::abs(delta_i);
if (i + delta_i >= 0 && i + delta_i < h
&& j + delta_j >= 0 && j + delta_j < w) {
ors.push_back((i + delta_i) * w + (j + delta_j));
}
if (delta_j > 0 && i + delta_i >= 0 && i + delta_i < h
&& j - delta_j >= 0 && j - delta_j < w) {
ors.push_back((i + delta_i) * w + (j - delta_j));
}
}
if (!ors.empty())
list.push_back(add_and({add_or(ors), i * w + j}));
}
}
add_or(list);
} else { // [S6]
vec ors;
for (int i = std::min(h - 1, K); i >= 0; i--) {
int j = K - i;
if (j >= w)
break;
ors.push_back(i * w + j);
}
add_or(ors);
}
}
