SDUT 2882 Full Binary Tree【思维】

最新推荐文章于 2018-04-11 19:51:02 发布

原创最新推荐文章于 2018-04-11 19:51:02 发布 · 409 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#SDUT 2882

思维专栏收录该内容

692 篇文章

订阅专栏

探讨无限全二叉树中两点间最短路径长度的计算方法，通过递归求解最近公共祖先来减少距离计算。

Full Binary Tree

Time Limit: 2000MS Memory Limit: 65536KB

Submit Statistic Discuss

Problem Description

In computer science, a binary tree is a tree data structure in which each node has at most two children. Consider an infinite full binary tree (each node has two children except the leaf nodes) defined as follows. For a node labelled v its left child will be labelled 2 * v and its right child will be labelled 2 * v + 1. The root is labelled as 1.

You are given n queries of the form i, j. For each query, you have to print the length of the shortest path between node labelled i and node labelled j.

Input

First line contains n(1 ≤ n ≤ 10^5), the number of queries. Each query consists of two space separated integers i and j(1 ≤ i, j ≤ 10^9) in one line.

Output

For each query, print the required answer in one line.

Example Input

5
1 2
2 3
4 3
1024 2048
3214567 9998877

Example Output

Hint

Author

2014年山东省第五届ACM大学生程序设计竞赛

题目大意：

二叉树遍历作为编号，问两个点之间的最短路。

思路：

ans=dist（1，i）+dist（1，j）-2*dist（1，Lca[i,j]）

找父亲节点的过程不断/2即可。

Ac代码：

#include <bits/stdc++.h>
using namespace std;
typedef long long int LL ;
const LL MOD = 1000000000+7;

#define abs(x)  (((x)>0)?(x):-(x))

/***************************************/
int path[500];
int path2[500];
int main(){
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int i,j;
        scanf("%d%d",&i,&j);
        int dist1=0;
        int dist2=0;
        while(i>1)
        {
            path[dist1]=i;
            i/=2;
            dist1++;
        }
        while(j>1)
        {
            path2[dist2]=j;
            j/=2;
            dist2++;
        }
        int flag=0;
        int pos=-1;
        for(int i=0;i<dist1;i++)
        {
            for(int j=0;j<dist2;j++)
            {
                if(path[i]==path2[j])
                {
                    pos=path[i];
                    flag=1;
                    break;
                }
            }
            if(flag==1)break;
        }
        int dist3=0;
        while(pos>1)
        {
            pos/=2;
            dist3++;
        }
        printf("%d\n",dist1+dist2-2*dist3);
    }
}