Submission #1308068

#	Time	Username	Problem	Language	Result	Execution time	Memory
1308068		orzorzorz	팀들 (IOI15_teams)	C++20	100 / 100	1834 ms	155744 KiB

teams

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

using namespace std;

// --- Persistent Segment Tree ---
// Allows querying: count students with B >= Y within a range of A values.
const int MAXN = 500005;
const int MAX_NODES = MAXN * 25;

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

int roots[MAXN]; // roots[i] stores the segment tree for students with A <= i
int node_cnt = 0;
int N_global;

// Build an empty tree
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: Add a student with upper bound 'pos' (B value)
int update(int prev_node, int tl, int tr, int pos) {
    int id = ++node_cnt;
    tree[id] = tree[prev_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 students with B in range [ql, qr]
// Since we access roots[R] - roots[L], we effectively filter by A first.
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);
}

// Counts students satisfying: min_size_pref > A_lower AND max_size_pref >= B_threshold
// This maps to: count students in PST range (A_lower, N] with value >= B_threshold
int count_students(int a_lower, int b_threshold) {
    if (b_threshold > N_global) return 0;
    // We want students with A in (a_lower, N_global].
    // Since roots[i] accumulates 1..i, roots[N] - roots[a_lower] gives range (a_lower, N].
    // Note: The problem logic often requires students with A <= K_i.
    // Let's stick to the DP recurrence: count(K_j < A <= K_i, B >= K_i).
    
    int cnt_all = query(roots[N_global], 1, N_global, b_threshold, N_global); // All with B >= K_i
    int cnt_low = query(roots[a_lower], 1, N_global, b_threshold, N_global);  // Those with A <= K_j
    
    // This function returns students with A > a_lower AND B >= b_threshold
    // But wait, the standard reduction usually uses: count(A <= a_upper, B >= b_threshold)
    // Let's adjust for the DP needs:
    // We usually compute: Count(A <= K_i, B >= K_i) - Count(A <= K_j, B >= K_i)
    return 0; // Placeholder, logic moved inside solve for clarity
}

// Specific helper for the DP transition
// Returns count of students with A <= a_limit AND B >= b_limit
int get_cnt(int a_limit, int b_limit) {
    if (b_limit > N_global) return 0;
    return query(roots[a_limit], 1, N_global, b_limit, N_global);
}

void init(int N, int A[], int B[]) {
    N_global = N;
    node_cnt = 0;
    
    vector<vector<int>> by_A(N + 1);
    for (int i = 0; i < N; i++) {
        by_A[A[i]].push_back(B[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 j
    int limit_y;    // The K value up to which this j is optimal
};

int can(int M, int K[]) {
    // Sort K to apply Hall's on intervals
    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());
    
    vector<long long> dp(M + 1); // dp[i] = max slack after satisfying 1..i
    dp[0] = 0;
    
    vector<StackItem> st;
    st.push_back({0, N_global + 1});
    
    for (int i = 1; i <= M; i++) {
        int k_curr = k_sorted[i];
        
        // 1. Maintain Convex Hull (Find best j)
        // We pop if the current requested size k_curr exceeds the validity of the top strategy
        while (st.size() > 1 && st.back().limit_y <= k_curr) {
            st.pop_back();
        }
        
        int j = st.back().idx;
        
        // 2. DP Recurrence
        // Slack[i] = Slack[j] + (Supply between j and i) - Demand[i]
        // Supply = Students with K[j] < A <= K[i] AND B >= K[i]
        // Calculated as: Count(A <= K[i], B >= K[i]) - Count(A <= K[j], B >= K[i])
        
        int count_upto_i = get_cnt(k_curr, k_curr);
        int count_upto_j = get_cnt(k_sorted[j], k_curr);
        
        dp[i] = dp[j] + (count_upto_i - count_upto_j) - k_curr;
        
        if (dp[i] < 0) return 0; // Hall's condition violated
        
        // 3. Update Convex Hull (Add i)
        // We need to find the split point 'ans' (a value of team size Y)
        // where strategy 'i' becomes worse than 'st.back()'.
        // Equation: dp[i] + Count(A<=Y, B>=Y) - Count(A<=K[i], B>=Y) >= dp[old] + Count(A<=Y, B>=Y) - Count(A<=K[old], B>=Y)
        // Simplifies to: dp[i] - Count(A<=K[i], B>=Y) >= dp[old] - Count(A<=K[old], B>=Y)
        // -> Count(A<=K[i], B>=Y) - Count(A<=K[old], B>=Y) <= dp[i] - dp[old]
        // LHS is monotonically non-increasing as Y increases (fewer students have B >= large Y)
        
        while (true) {
            int old_idx = st.back().idx;
            long long diff_dp = dp[i] - dp[old_idx];
            
            // Binary Search for the split point
            int low = k_curr + 1;
            int high = N_global + 1;
            int ans = N_global + 1;
            
            while (low <= high) {
                int mid = (low + high) / 2;
                // Cost function difference
                int val = get_cnt(k_curr, mid) - get_cnt(k_sorted[old_idx], mid);
                
                if (val <= diff_dp) {
                    ans = mid;
                    high = mid - 1;
                } else {
                    low = mid + 1;
                }
            }
            
            if (ans >= st.back().limit_y) {
                st.pop_back();
                if (st.empty()) {
                    st.push_back({i, N_global + 1});
                    break;
                }
            } else {
                st.push_back({i, ans});
                break;
            }
        }
    }
    
    return 1;
}

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