Submission #1018788

#	Time	Username	Problem	Language	Result	Execution time	Memory
1018788		ProtonDecay314	Mechanical Doll (IOI18_doll)	C++17	53 / 100	81 ms	14784 KiB

doll

// AM+DG

/*

*/
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef vector<ll> vll;
typedef vector<vll> vvll;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef pair<int, int> pi;
typedef pair<ll, ll> pll;
typedef vector<pi> vpi;
typedef vector<pll> vpll;
#define L(i, varmn, varmx) for(ll i = varmn; i < varmx; i++)
#define LR(i, varmx, varmn) for(ll i = varmx; i > varmn; i--)
#define LI(i, varmn, varmx) for(int i = varmn; i < varmx; i++)
#define LIR(i, varmx, varmn) for(int i = varmx; i > varmn; i--)
#define pb push_back
#include "doll.h"

struct pot_switch {
    int id;
    int x;
    int y;
};

void construct_switches(int base_switch_num, const vi& to_connect, int p2, int offset, int size, vector<pot_switch>& switches_to_construct) {

    if(size == 2) {
        // Handle the special case here
        switches_to_construct.pb({base_switch_num - offset, to_connect[p2 + offset - ((p2 << 1) - 1)], to_connect[(p2 << 1) + offset - ((p2 << 1) - 1)]});
    } else if(size == 3) {
        // Handle the special case here
        switches_to_construct.pb({base_switch_num - offset, base_switch_num - (offset + p2), to_connect[(p2 << 1) + offset - ((p2 << 1) - 1)]});

        // Recurse towards the first child
        construct_switches(base_switch_num, to_connect, p2 << 1, offset + p2, size - (size >> 1), switches_to_construct);
    } else {
        // Recursive case
        switches_to_construct.pb({base_switch_num - offset, base_switch_num - (offset + p2), base_switch_num - (offset + (p2 << 1))});

        // Recurse left
        construct_switches(base_switch_num, to_connect, p2 << 1, offset + p2, size - (size >> 1), switches_to_construct);

        // Recurse right
        construct_switches(base_switch_num, to_connect, p2 << 1, offset + (p2 << 1), (size >> 1), switches_to_construct);
    }
}

int ceillog2(int n) {
    int p2 = 1;

    while(p2 < n) {
        p2 <<= 1;
    }

    return p2;
}

void create_circuit(int m, std::vector<int> a) {
    int n = a.size();
    vi c(m + 1);
    c[0] = -1;
    for (int i = 1; i <= m; ++i) {
        c[i] = 1;
    }
    // vi x(N), y(N);
    // for (int k = 0; k < N; ++k) {
    //     x[k] = y[k] = a[k];
    // }

    vi x;
    vi y;

    vvi adjmat;

    LI(i, 0, m + 1) {
        vi adjmatr;
        adjmat.pb(adjmatr);
    }

    LI(i, -1, n) {
        if(i == -1) {
            adjmat[0].pb(a[i + 1]);
        } else if(i == n - 1) {
            adjmat[a[i]].pb(0);
        } else {
            adjmat[a[i]].pb(a[i + 1]);
        }
    }

    int last_switch = 0;
    LI(i, 0, m + 1) {
        int cur_num_next = adjmat[i].size();
        if(cur_num_next == 0) {
            c[i] = 0;
        } else if(cur_num_next == 1) {
            c[i] = adjmat[i][0];
        } else {
            /*
            Construct switches needs...
            whatever the last switch was,
            the list of things it should connect the leaf node to
            the current power of two separation (starts at 1 and doubles)
            the current offset (starts at 0)
            for the left tree, it increments by power * 1
            for the right, its power * 2
            Left tree must be of size ceil(N / 2)
            right is size floor(N / 2)

            */

            c[i] = last_switch - 1;

            int actual_num_next = ceillog2(adjmat[i].size());
            vi actual_next;
            actual_next.reserve(actual_num_next);
            LI(j, 0, actual_num_next - cur_num_next) {
                actual_next.pb(last_switch - 1);
            }
            LI(j, 0, cur_num_next) {
                actual_next.pb(adjmat[i][j]);
            }

            vector<pot_switch> switches_to_construct;
            construct_switches(last_switch - 1, actual_next, 1, 0, actual_num_next, switches_to_construct);
            sort(switches_to_construct.begin(), switches_to_construct.end(), [](pot_switch s1, pot_switch s2) {return s1.id > s2.id;}); // Sort by decreasing ID
            LI(j, 0, actual_num_next - 1) {
                x.pb(switches_to_construct[j].x);
                y.pb(switches_to_construct[j].y);
            }
            last_switch -= actual_num_next - 1;
        }
    }

    answer(c, x, y);

    return;
}

• Subtask 1: 2 / 2
• Subtask 2: 4 / 4
• Subtask 3: 10 / 10
• Subtask 4: 0 / 10
• Subtask 5: 9.0000000036 / 18
• Subtask 6: 28.0000000112 / 56