Submission #1321761

#	Time	Username	Problem	Language	Result	Execution time	Memory
1321761		benjaminkleyn	이주 (IOI25_migrations)	C++20	0 / 100	39 ms	1192 KiB

migrations

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

int n;
int P[10000];
vector<int> g[10000];
int dist[10000] = {0};
void dfs(int u, int p = -1)
{
    if (p == -1) dist[u] = 0;
    for (int v : g[u]) if (v != p) {
        dist[v] = dist[u] + 1;
        dfs(v, u);
    }
}

int prev_a, prev_b;
int a, b;
void find_diameter()
{
    prev_a = a;
    prev_b = b;
    a = b = 0;
    dfs(0);
    for (int i = 0; i < n; i++)
        if (dist[i] > dist[a])
            a = i;

    dfs(a);
    for (int i = 0; i < n; i++)
        if (dist[i] > dist[b])
            b = i;

    if (a > b) swap(a, b);
}

bool cancelled = false;
int send_message(int N, int i, int Pi)
{
    n = N;
    g[i].push_back(Pi);
    g[Pi].push_back(i);

    if (N - 38 <= i && i < N - 24) {
        find_diameter();
        return (a >> (2 * (i - (N - 38)))) & 0b11;
    }
    if (N - 24 <= i && i < N - 10) {
        find_diameter();
        return (b >> (2 * (i - (N - 24)))) & 0b11;
    }
    if (N - 10 <= i && i < N - 5) {
        find_diameter();
        if (a == i)
            return 3;
        if (b == i)
            return 4;
        if (N - 38 <= a && a < N - 10)
            return ((N - 10 - a) >> (i - (N - 10))) & 1;
        return 0;
    }
    if (N - 5 <= i) {
        find_diameter();
        if (a == i)
            return 3;
        if (b == i)
            return 4;
        if (N - 38 <= b && b < N - 10)
            return ((N - 10 - b) >> (i - (N - 5))) & 1;
        return 0;
    }
    return 0;
}

pair<int, int> longest_path(vector<int> S)
{
    int N = S.size();
    int a = 0, b = 0;

    for (int i = N - 38; i < N - 24; i++)
        a |= S[i] << (2 * (i - (N - 40)));
    for (int i = N - 24; i < N - 10; i++)
        b |= S[i] << (2 * (i - (N - 24)));

    int offset_a = 0, offset_b = 0;
    for (int i = N - 10; i < N - 5; i++)
        offset_a |= S[i] << (i - (N - 10));
    for (int i = N - 5; i < N; i++)
        offset_b |= S[i] << (i - (N - 5));

    if (0 < offset_a && offset_a <= 28)
        a = N - 10 - offset_a;
    if (0 < offset_b && offset_b <= 28)
        b = N - 10 - offset_b;

    for (int i = N - 10; i < N; i++) {
        if (S[i] == 3)
            a = i;
        if (S[i] == 4)
            b = i;
    }

    return {a, b};
}

• Subtask 0: 0 / 0
• Subtask 1: 0 / 30
• Subtask 2: 0 / 70