HDU 5323 Solve this interesting problem（DFS）_g - solve this interesting problem-优快云博客

本文链接：https://blog.youkuaiyun.com/yeguxin/article/details/47129049

本博客探讨了如何通过区间查询找到满足特定条件的最小整数n，包括问题描述、输入输出格式、示例解析及解决方案。采用深度优先搜索（DFS）策略遍历段树，以求解区间[L, R]的最小满足条件的整数。

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Solve this interesting problem

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1507 Accepted Submission(s): 428

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

Sample Output

Source

2015 Multi-University Training Contest 3

题意：求满足输入区间的最小整数n，没有的话输出-1

思路：四种情况，直接爆搜即可

#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<string.h>
#include<stdlib.h>

using namespace std;

int n,m;
int minn;

void DFS(int l,int r)
{
    if(minn && r>=minn)
    {
        return ;
    }

    if(l == 0)
    {
        if(minn)
        {
            minn = min(r,minn);
        }
        else
        {
            minn = r;
        }
        return ;
    }
    if((l+l)<(r+1))
    {
        return ;
    }
    
    DFS((l-1)*2-r,r);
    DFS((l-1)*2+1-r,r);
    DFS(l,2*r-l);
    DFS(l,2*r-l+1);

}

int main()
{
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        minn = 0;
        if(m == 0)
        {
            printf("0\n");
            continue;
        }
        DFS(n,m);
        if(minn)
        {
            printf("%d\n",minn);
        }
        else
        {
            printf("-1\n");
        }
    }
    return 0;
}