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Efficient Algorithms: Binary Search on Answer and Divide-and-Conquer Structures

Published at: 2026-05-27 09:17 Last updated: 2026-06-06 13:04
#algorithm #Divide and Conquer #Binary Search

Core Logic and Mathematical Principles

Binary Search on Answer

Monotonicity is the mathematical cornerstone of binary search on the answer. Let the solution space be $D$ and the decision function be $f: D \to \{0, 1\}$. If $f(x)$ satisfies the property of monotonicity (either increasing or decreasing), that is:

$$\forall x_1, x_2 \in D, x_1 < x_2 \implies f(x_1) \le f(x_2)$$

then we can approach the optimal solution $x^*$ by performing a binary search on the solution space $D$ within $O(\log |D|)$ evaluations. Each decision eliminates half of the invalid value range, transforming the optimization problem into a decision problem.

Divide and Conquer Tree Structure

Transform the spatial intervals of the sequence $A[1..N]$ into a temporal tree-based divide-and-conquer process. For the interval $[l, r]$, take $mid = \lfloor \frac{l+r}{2} \rfloor$, recursively constructing the left subtree $[l, mid]$ and the right subtree $[mid+1, r]$. The entire divide-and-conquer trajectory is logically abstracted into a complete binary tree of height $O(\log N)$. By merging this logical binary tree from the bottom up (as in the PushUp logic of segment trees) or transmitting from the top down (like the left contribution to the right in CDQ divide-and-conquer), we can decompose the $O(N^2)$ global sequence problem into the summation of $\log N$ layers of local intervals. The total workload at each layer of the tree is controlled within $O(N)$, resulting in an overall time complexity converging to $O(N \log N)$.

Algorithm Derivation and Divide-and-Conquer Structure Design

Boundary Derivation of Binary Search on Answer

For decision functions satisfying the monotonicity of "all $1$s followed by all $0$s", we seek the largest value $x^*$ such that $f(x)=1$. Initialize two pointers $l = \min(D), r = \max(D)$.

  1. Set $mid = l + \lfloor \frac{r-l+1}{2} \rfloor$ to prevent integer overflow caused by $l+r$, and use ceiling to avoid infinite loops when $r = l+1$.
  2. If $f(mid) == 1$, it indicates that the answer lies within the closed interval $[mid, r]$; update $l = mid$.
  3. If $f(mid) == 0$, it indicates that the answer lies within the closed interval $[l, mid-1]$; update $r = mid - 1$.
  4. Terminate when $l == r$, at which point $l$ is the desired solution.

Interval Decomposition of Divide-and-Conquer Tree

In sequence divide-and-conquer, any query interval $[L, R]$ will be accurately cut into $O(\log N)$ non-leaf nodes on the divide-and-conquer binary tree. Let the current divide-and-conquer node be $u$, governing the sequence interval $[l, r]$.

  1. If $[L, R] \supseteq [l, r]$, directly extract the preprocessed interval features of that node and terminate recursion.
  2. If $L \le mid$, recursively enter the left subtree $(u \ll 1)$.
  3. If $R > mid$, recursively enter the right subtree $(u \ll 1 | 1)$. The statistical crossing over $mid$ fundamentally involves the algebraic convolution of the suffix information from the left subtree and the prefix information from the right subtree.

C++ Standard Source Code

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

using namespace std;

// Simulate classic NOIP problem: maximizing the minimum value problem (e.g., cutting sticks, jumping stones)
// Decision function: checks if the sequence can be divided into at least k segments under the constraint that each segment has at least length check_val
bool check_feasibility(const vector<int>& stone_pos, int target_segments, int max_dist, int check_val) {
    int count = 0;
    int last_pos = 0;

    for (size_t i = 0; i < stone_pos.size(); ++i) {
        if (stone_pos[i] - last_pos >= check_val) {
            count++;
            last_pos = stone_pos[i];
        }
    }
    // Critical pitfall: the decision boundary must strictly contain the last segment distance to the endpoint, otherwise, it leads to logical errors
    if (max_dist - last_pos < check_val) {
        count--;
    }

    return count >= target_segments;
}

// Divide-and-Conquer Tree Structure: Maximum Continuous Subarray Sum (simulating core logic of segment tree)
struct SegmentNode {
    long long sum;
    long long lmax;
    long long rmax;
    long long tmax;
};

// Bottom-up merge logic
SegmentNode merge_nodes(const SegmentNode& left, const SegmentNode& right) {
    SegmentNode res;
    res.sum = left.sum + right.sum;
    res.lmax = max(left.lmax, left.sum + right.lmax);
    res.rmax = max(right.rmax, right.sum + left.rmax);
    // Core logic for crossing mid: left subtree's right max + right subtree's left max
    res.tmax = max({left.tmax, right.tmax, left.rmax + right.lmax});
    return res;
}

// Building process for the divide-and-conquer tree
void build_divide_tree(vector<SegmentNode>& tree, const vector<int>& a, int u, int l, int r) {
    if (l == r) {
        tree[u] = {a[l], a[l], a[l], a[l]};
        return;
    }
    int mid = l + ((r - l) >> 1); // Critical pitfall: bit operation precedence is low, must add outer parentheses
    build_divide_tree(tree, a, u << 1, l, mid);
    build_divide_tree(tree, a, u << 1 | 1, mid + 1, r);
    tree[u] = merge_nodes(tree[u << 1], tree[u << 1 | 1]);
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

    // 1. Testing the binary search module
    int L = 25, n = 5, m = 2; // Total length 25, 5 stones, remove 2 stones (i.e., divide into 4 segments)
    vector<int> stones = {2, 11, 14, 17, 21};

    int l_ans = 1, r_ans = L, ans = 0;
    while (l_ans <= r_ans) {
        int mid_ans = l_ans + ((r_ans - l_ans) >> 1);
        if (check_feasibility(stones, n - m, L, mid_ans)) {
            ans = mid_ans;
            l_ans = mid_ans + 1; // Searching for the maximum value, approaching the right
        } else {
            r_ans = mid_ans - 1;
        }
    }

    // 2. Testing the divide tree module
    int seq_len = 4;
    vector<int> seq = {0, -1, 2, 3}; // 1-indexed simulation
    vector<SegmentNode> divide_tree(seq_len * 4);
    build_divide_tree(divide_tree, seq, 1, 1, seq_len - 1);

    return 0;
}

NOIP Practical Pitfall Guide

  1. Infinite loops and incorrect divisions caused by bit operation precedence To pursue constant optimization, contestants often write mid = (l + r) / 2 as mid = l + r >> 1 or mid = l + (r - l) >> 1. Due to the higher precedence of addition and subtraction operators + and - over the bitwise shift operator >>, the expression will be interpreted as mid = (l + (r - l)) >> 1 or even mid = (l + r) >> 1. When $l$ and $r$ are large, this can lead to integer overflow or cause infinite loops due to incorrect calculations during binary search boundary contraction. Parentheses must be enforced: mid = l + ((r - l) >> 1).
  2. Infinite loops and value domain overflow vulnerabilities in binary search When writing different binary searches for "maximizing the minimum value" and "minimizing the maximum value", contestants often confuse pointer updates. If mid is calculated using left rounding l + ((r - l) >> 1), when l = r - 1, mid will always equal l. If the branch logic then sets l = mid, the interval will no longer shrink, leading to an infinite loop. It is crucial to strictly correspond the adjustment of mid's rounding direction based on whether l = mid or r = mid.

Classic NOIP/Luogu Problems

Luogu P2678 [NOIP2015 Advanced Group] Jumping Stones

Luogu P4145 The God Creates Questions in Seven Minutes 2 / The Flower Goddess Travels the World

Original Source: local://3.2
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