HUD 5323 Solve this interesting problem dfs剪枝搜索，重构线段树

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本文介绍了一个有趣的算法问题：如何找到一个最小的非负整数n，使得存在一个根节点范围为[0,n]的线段树能够完全覆盖给定的[L,R]区间。通过递归地搜索所有可能的子区间来解决这个问题，并实现了一个高效的C++程序。

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Solve this interesting problem

Problem Description

Have you learned something about segment tree? If not, don’t worry, I will explain it for you.
Segment Tree is a kind of binary tree, it can be defined as this:
- For each node u in Segment Tree, u has two values:

Lu and

Ru.
- If

Lu=Ru, u is a leaf node.
- If

Lu≠Ru, u has two children x and y,with

Lx=Lu,

Rx=⌊Lu+Ru2⌋,

Ly=⌊Lu+Ru2⌋+1,

Ry=Ru.
Here is an example of segment tree to do range query of sum.

Given two integers L and R, Your task is to find the minimum non-negative n satisfy that: A Segment Tree with root node's value

Lroot=0 and

Rroot=n contains a node u with

Lu=L and

Ru=R.

Input

The input consists of several test cases.
Each test case contains two integers L and R, as described above.

0≤L≤R≤109

LR−L+1≤2015

Output

For each test, output one line contains one integer. If there is no such n, just output -1.

Sample Input

6 7
10 13
10 11

Sample Output

7
-1
12

/*
 * 给定一个区间的 L R 问知否存在这样的线段树 [0 , n] 使得n最小并且包含区间 [L, R]
 * dfs向上搜索， 由于该区间可能是父节点的左儿子或右儿子，分类讨论，注意取整时可能
 * 导致求得的L 或 R 与实际相差1 ， 将这四种情况都考虑即可。
 * 注意剪枝， 在线段树中，每个节点的左儿子区间 = 右儿子区间 或者 右儿子加一
 */

#include <bits/stdc++.h>
#define endl "\n"
using namespace std;
typedef long long ll;

ll ans = -1;

void dfs(ll l, ll r) {
    if(l == 0) {
        if(ans == -1) {
            ans = r;
        }
        else {
            ans = min(ans, r);
        }
        return ;
    }
    if(l < 0) return ;
    if(ans != -1 && r >= ans) return ;
    if(l < r-l+1) return; // 剪枝
    dfs(l, 2*r - l);
    dfs(l, 2*r - l + 1);
    dfs(2*l - r - 1, r);
    dfs(2*l - r - 2, r);
}

int main() {
    ios::sync_with_stdio(false);
    ll l, r;
    while(cin >> l >> r) {
        if(l == r) {
            cout << l << endl;
            continue;
        }
        ans = -1;
        dfs(l, r);
        cout << ans << endl;
    }
    return 0;
}