Submission #1352406

#TimeUsernameProblemLanguageResultExecution timeMemory
1352406tickcrossyMigrations (IOI25_migrations)C++20
30 / 100
22 ms972 KiB
#include <vector>
#include <algorithm>

using namespace std;

// 倍增LCA及树深维护
static int depth[10005];
static int up[14][10005];

// 在线直径和内部状态维护
static int U_node = 0, V_node = 0, D_tree = 0;
static long long id_val = 0;
static int bit_idx = 0;

// O(log N) 查询两点距离
int get_lca(int u, int v) {
    if (depth[u] < depth[v]) swap(u, v);
    int diff = depth[u] - depth[v];
    for (int j = 0; j < 14; j++) {
        if ((diff >> j) & 1) u = up[j][u];
    }
    if (u == v) return u;
    for (int j = 13; j >= 0; j--) {
        if (up[j][u] != up[j][v]) {
            u = up[j][u];
            v = up[j][v];
        }
    }
    return up[0][u];
}

int get_dist(int u, int v) {
    return depth[u] + depth[v] - 2 * depth[get_lca(u, v)];
}

// ================== 研究团队 ==================
int send_message(int N, int i, int Pi) {
    // 复位内部所有状态防评测机多次复用
    if (i == 1) { 
        depth[0] = 0;
        for (int j = 0; j < 14; j++) up[j][0] = 0;
        U_node = 0; V_node = 0; D_tree = 0;
        id_val = 0; bit_idx = 0;
    }

    // 更新倍增表
    depth[i] = depth[Pi] + 1;
    up[0][i] = Pi;
    for (int j = 1; j < 14; j++) {
        up[j][i] = up[j-1][up[j-1][i]];
    }

    int dU = get_dist(i, U_node);
    int dV = get_dist(i, V_node);
    bool increased = false;
    int replaced = 0; 
    int max_d = max(dU, dV);
    
    // 动态直径贪心更迭机制
    if (max_d > D_tree) {
        increased = true;
        D_tree = max_d;
        if (dU > dV) {
            replaced = 1; 
            V_node = i;
        } else if (dV > dU) {
            replaced = 2; 
            U_node = i;
        } else { // dU == dV,绝对偏袒保留 0
            if (U_node == 0) {
                replaced = 1; 
                V_node = i;
            } else if (V_node == 0) {
                replaced = 2; 
                U_node = i;
            } else {
                replaced = 1; 
                V_node = i;
            }
        }
    }

    // 在 N-50 步,冻结进入密文模式前最终始的基础直径端点身份
    if (i == N - 50) { 
        // 将包含 0 的端点强制交换至 U_node
        if (V_node == 0) {
            swap(U_node, V_node);
            if (increased) replaced = 3 - replaced; 
        }
        // U 位于低 14 位,V 位于高 14 位
        id_val = V_node * 16384LL + U_node; 
        bit_idx = 0;
    }

    // ====== 传输层 (M <= 50) ======
    if (i >= N - 50) {
        if (increased) {
            // 事件抢占帧:3 代表 V 被更新,4 代表 U 被更新
            return replaced == 1 ? 3 : 4; 
        } else {
            // 平静期发送数据帧
            if (bit_idx < 28) {
                long long b = (id_val >> bit_idx) & 1;
                bit_idx++;
                return b + 1; // 发送 1(对于位0) 或 2(对于位1)
            }
        }
    }

    return 0; // 前 9950 步保持彻底静默,规避通信代价 M
}

// ================== 博物馆 ==================
std::pair<int, int> longest_path(std::vector<int> S) {
    int N = S.size();
    long long id = 0;
    int final_U = -1, final_V = -1;
    int b_idx = 0;
    
    // 从最后 50 步的信号列恢复真相
    for (int i = N - 50; i < N; i++) {
        if (i < 0) continue; // 边界保护
        int val = S[i];
        if (val == 0) continue; 
        
        if (val == 3) {
            final_V = i;  // 更新事件信号
        } else if (val == 4) {
            final_U = i;  // 更新事件信号
        } else if (val == 1 || val == 2) {
            if (b_idx < 28) { // 组装基础数据位
                id |= ((long long)(val - 1) << b_idx);
                b_idx++;
            }
        }
    }
    
    // 解析底座 (低 14 位 为 U,高 14 位 为 V)
    int initial_U = id % 16384;
    int initial_V = id / 16384;
    
    // 如果无抢占事件,沿用底座恢复结果
    if (final_U == -1) final_U = initial_U;
    if (final_V == -1) final_V = initial_V;
    
    return {final_U, final_V};
}
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