Submission #1321783

#	Time	Username	Problem	Language	Result	Execution time	Memory
1321783		benjaminkleyn	Migrations (IOI25_migrations)	C++20	72.89 / 100	49 ms	1448 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;
    int max_dist = dist[b];

    // can we keep one of the two the same?
    dfs(prev_a);
    for (int i = 0; i < n; i++)
        if (dist[i] == max_dist) {
            a = prev_a, b = i;
            return;
        }

    dfs(prev_b);
    for (int i = 0; i < n; i++)
        if (dist[i] == max_dist) {
            b = prev_b, a = i;
            return;
        }
}

// can both change at once?
// can it go from (u,v) -> (x,y) where none of u,v,x,y are equal?
// well then surely the intermediate step (u,x) or (v,x) must have been answer?


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

    if (N - 28 <= i && i < N - 14) {
        find_diameter();
        int send = cancelled_a ? 0 : ((a >> (i - (N - 28))) & 0b1);
        if (a == i) {
            send = 4;
            cancelled_a = true;
        }
        if (b == i) {
            send |= 2;
            cancelled_b = true;
        }
        return send;
    }
    if (N - 14 <= i && i < N) {
        find_diameter();
        int send = cancelled_b ? 0 : (b >> (i - (N - 14))) & 0b1;
        if (a == i) {
            send |= 2;
            cancelled_a = true;
        }
        if (b == i) {
            send = 4;
            cancelled_b = true;
        }
        return send;
    }
    return 0;
}

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

    for (int i = N - 28; i < N - 14; i++)
        a |= (S[i] & 1) << (i - (N - 28));
    for (int i = N - 14; i < N; i++)
        b |= (S[i] & 1) << (i - (N - 14));

    for (int i = N - 28; i < N - 14; i++) {
        if (S[i] == 4)
            a = i;
        if (S[i] & 2)
            b = i;
    }
    for (int i = N - 14; i < N; i++) {
        if (S[i] == 4)
            b = i;
        if (S[i] & 2)
            a = i;
    }

    return {a, b};
}

• Subtask 0: 0 / 0
• Subtask 1: 30 / 30
• Subtask 2: 42.8897 / 70