제출 #603777

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
603777		jophyyjh	Vision Program (IOI19_vision)	C++14	33 / 100	97 ms	1104 KiB

vision

/**
 * 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     (i think. i haven't checked)
 * Number of reads:	6HW
 * Implementation 1         (Solves [S1-3], [S5], [S6-7])
*/

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

typedef std::vector<int>    vec;


void construct_network(int h, int w, int K) {
    if (K == 1 || h * w <= 900) {
        vec ands;
        for (int i1 = 0; i1 < h; i1++) {
            for (int j1 = 0; j1 < w; j1++) {
                for (int i2 = i1; i2 < h; i2++) {
                    for (int j2 = 0; j2 < w; j2++) {
                        if (std::abs(i2 - i1) + std::abs(j2 - j1) == K)
                            ands.push_back(add_and({i1 * w + j1, i2 * w + j2}));
                    }
                }
            }
        }
        add_or(ands);
    } else {    // [S6]
        vec ands;
        for (int i = std::min(h - 1, K); i >= 0; i--) {
            int j = K - i;
            if (j >= w)
                break;
            ands.push_back(add_and({0, i * w + j}));
        }
        add_or(ands);
    }
}

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