Submission #657863

#	Time	Username	Problem	Language	Result	Execution time	Memory
657863		benjaminkleyn	Vision Program (IOI19_vision)	C++17	12 / 100	34 ms	4040 KiB

vision

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

void construct_network(int h, int w, int k)
{
    vector<int> rd1(h + w - 1), rd2(h + w - 1);
    vector<int> ld1(h + w - 1), ld2(h + w - 1);
    for (int d = 0; d < h + w - 1; d++)
    {
        // right-diagonal of (i, j) = i + j
        vector<int> idx;
        for (int i = 0; i < h; i++)
            if (0 <= d - i && d - i < w)
                idx.push_back(i * w + d - i);
        // rd1 : is there at least one black pixel on this diagonal?
        rd1[d] = add_or(idx);
        // rd2 : are there exactly two black pixels on this diagonal?
        rd2[d] = add_xor({add_xor(idx), rd1[d]});

        // left-diagonal of (i, j) = i - j + (w - 1)
        idx.clear();
        for (int i = 0; i < h; i++)
            if (0 <= i - d + w - 1 && i - d + w - 1 < w)
                idx.push_back(i * w + i - d + w - 1);
        // ld1 : is there at least one black pixel on this diagonal?
        ld1[d] = add_or(idx);
        // ld2 : are there exactly two black pixels on this diagonal?
        ld2[d] = add_xor({add_xor(idx), rd1[d]});
    }
    vector<int> A, B, C, D;
    for (int d = 0; d + k < h + w - 1; d++)
    {
        vector<int> idxR1, idxR2, idxL1, idxL2;
        for (int i = 0; i < k + 1; i++)
        {
            idxR1.push_back(rd1[d + i]);
            idxR2.push_back(rd2[d + i]);
            idxL1.push_back(ld1[d + i]);
            idxL2.push_back(ld2[d + i]);
        }
        A.push_back(add_or({add_xor({add_or(idxR1), add_xor(idxR1)}),add_or(idxR2)}));
        B.push_back(add_or({add_xor({add_or(idxL1), add_xor(idxL1)}),add_or(idxL2)}));
    }

    for (int d = 0; d + k - 1 < h + w - 1; d++)
    {
        vector<int> idxR1, idxR2, idxL1, idxL2;
        for (int i = 0; i < k; i++)
        {
            idxR1.push_back(rd1[d + i]);
            idxR2.push_back(rd2[d + i]);
            idxL1.push_back(ld1[d + i]);
            idxL2.push_back(ld2[d + i]);
        }
        C.push_back(add_or({add_xor({add_or(idxR1), add_xor(idxR1)}),add_or(idxR2)}));
        D.push_back(add_or({add_xor({add_or(idxL1), add_xor(idxL1)}),add_or(idxL2)}));
    }
    int x = add_and({add_or(A), add_or(B)});
    int y = add_and({add_or(C), add_or(D)});

    add_and({x, add_not(y)});
}

• Subtask 1: 0 / 10
• Subtask 2: 0 / 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