Submission #1308065

#	Time	Username	Problem	Language	Result	Execution time	Memory
1308065		orzorzorz	Teams (IOI15_teams)	C++20	100 / 100	1849 ms	156360 KiB

teams

#include <vector>
#include <algorithm>
#include <iostream>

using namespace std;

// --- Persistent Segment Tree Settings ---
const int MAXN = 500005;
const int MAX_NODES = MAXN * 25; // Sufficient buffer for N log N updates

struct Node {
    int l, r;
    int sum;
} tree[MAX_NODES];

int roots[MAXN]; // Roots for PST versions: roots[i] stores state for students with A <= i
int node_cnt = 0;
int N_global;

// --- PST Functions ---

int build(int tl, int tr) {
    int id = ++node_cnt;
    tree[id].sum = 0;
    tree[id].l = 0;
    tree[id].r = 0;
    if (tl != tr) {
        int tm = (tl + tr) / 2;
        tree[id].l = build(tl, tm);
        tree[id].r = build(tm + 1, tr);
    }
    return id;
}

// Update: increment count at position 'pos' (which corresponds to a B-value)
int update(int prev_node, int tl, int tr, int pos) {
    int id = ++node_cnt;
    tree[id] = tree[prev_node]; // Copy previous node
    tree[id].sum++;
    
    if (tl == tr) return id;
    
    int tm = (tl + tr) / 2;
    if (pos <= tm)
        tree[id].l = update(tree[prev_node].l, tl, tm, pos);
    else
        tree[id].r = update(tree[prev_node].r, tm + 1, tr, pos);
        
    return id;
}

// Query: count number of students with B-value in [ql, qr]
int query(int node, int tl, int tr, int ql, int qr) {
    if (ql > qr) return 0;
    if (ql == tl && qr == tr) return tree[node].sum;
    
    int tm = (tl + tr) / 2;
    if (qr <= tm) return query(tree[node].l, tl, tm, ql, qr);
    if (ql > tm) return query(tree[node].r, tm + 1, tr, ql, qr);
    
    return query(tree[node].l, tl, tm, ql, tm) + 
           query(tree[node].r, tm + 1, tr, tm + 1, qr);
}

// Helper: Count students with A <= a_lim AND B >= b_lim
int count_students(int a_lim, int b_lim) {
    if (b_lim > N_global) return 0;
    return query(roots[a_lim], 1, N_global, b_lim, N_global);
}

// --- Main Interface Functions ---

void init(int N, int A[], int B[]) {
    N_global = N;
    node_cnt = 0;
    
    // 1. Group students by their minimum team size A[i]
    // using vector of vectors to handle duplicate A values easily
    vector<vector<int>> by_A(N + 1);
    for (int i = 0; i < N; i++) {
        by_A[A[i]].push_back(B[i]);
    }
    
    // 2. Build Persistent Segment Tree
    // roots[i] represents the state including all students with A <= i
    roots[0] = build(1, N);
    
    for (int i = 1; i <= N; i++) {
        roots[i] = roots[i-1];
        for (int b_val : by_A[i]) {
            roots[i] = update(roots[i], 1, N, b_val);
        }
    }
}

struct StackItem {
    int idx;        // Index of the group in the sorted K array
    int limit_y;    // The B-value limit where this strategy stops being optimal
};

int can(int M, int K[]) {
    // 1. Prepare and Sort K
    // We use 1-based indexing for logic, k_sorted[0] is a dummy 0.
    vector<int> k_sorted(M + 1);
    k_sorted[0] = 0;
    for (int i = 0; i < M; i++) k_sorted[i+1] = K[i];
    sort(k_sorted.begin() + 1, k_sorted.end());
    
    // dp[i] stores the minimum "slack" (surplus capacity) after satisfying groups 1..i
    vector<long long> dp(M + 1);
    dp[0] = 0;
    
    // Stack for Convex Hull / Envelope
    // Stores indices j that form the lower envelope of cost functions.
    // 'Newer' items (higher j) are better for small Y (team sizes).
    // 'Older' items are better for large Y.
    vector<StackItem> st;
    st.push_back({0, N_global + 1}); // Base case: Group 0 valid for all sizes initially
    
    for (int i = 1; i <= M; i++) {
        int k_curr = k_sorted[i];
        
        // 2. Query Phase: Determine optimal previous state 'j'
        // Pop elements from the stack that are no longer optimal for the current team size k_curr.
        // We need the strategy whose range COVERS k_curr.
        // Stack item is valid for range [..., limit_y). 
        // If limit_y <= k_curr, the validity range has ended.
        while (st.size() > 1 && st.back().limit_y <= k_curr) {
            st.pop_back();
        }
        
        int j = st.back().idx;
        
        // 3. Calculate DP
        // Recurrence: dp[i] = dp[j] + count(K[j] < A <= K[i], B >= K[i]) - K[i]
        int cnt_i = count_students(k_curr, k_curr);
        int cnt_j = count_students(k_sorted[j], k_curr);
        
        dp[i] = dp[j] + (cnt_i - cnt_j) - k_curr;
        
        // If slack is negative, it's impossible to satisfy demands
        if (dp[i] < 0) return 0;
        
        // 4. Update Phase: Add 'i' to the stack
        // We need to find the split point 'ans' such that:
        // For y < ans, 'i' is better. For y >= ans, 'st.back()' (Old) is better.
        // Condition for Old Better: dp[i] - count(K[i], y) >= dp[old] - count(K[old], y)
        // -> count(K[i], y) - count(K[old], y) <= dp[i] - dp[old]
        
        while (true) {
            int old_idx = st.back().idx;
            long long diff_dp = dp[i] - dp[old_idx];
            
            // Binary search for the first point where Old becomes better (or equal)
            int low = k_curr + 1;
            int high = N_global + 1;
            int ans = N_global + 1; // Default: Old is never better
            
            while (low <= high) {
                int mid = (low + high) / 2;
                int count_diff = count_students(k_curr, mid) - count_students(k_sorted[old_idx], mid);
                
                if (count_diff <= diff_dp) {
                    ans = mid;
                    high = mid - 1; // Try to find an earlier split point
                } else {
                    low = mid + 1;
                }
            }
            
            // If the calculated split point 'ans' is beyond or equal to the limit of the current Old,
            // then Old is completely overshadowed by New (or Old's validity interval is empty).
            // We pop Old and try comparing 'i' against the next element in stack.
            if (ans >= st.back().limit_y) {
                st.pop_back();
                if (st.empty()) {
                    // This case handles when 'i' beats the base case (should essentially never happen 
                    // unless i is better for all Y up to N, in which case it becomes new base)
                    st.push_back({i, N_global + 1});
                    break;
                }
            } else {
                // Found a valid split. Old is restricted to valid up to 'ans', New takes over before that.
                // Wait, stack structure: Top is New, below is Old.
                // Top's limit is defined by when Old takes over.
                st.push_back({i, ans});
                break;
            }
        }
    }
    
    return 1;
}

• Subtask 1: 21 / 21
• Subtask 2: 13 / 13
• Subtask 3: 43 / 43
• Subtask 4: 23 / 23