Problem 041—— UVa 679 - Dropping Balls

最新推荐文章于 2020-09-30 23:26:56 发布

原创最新推荐文章于 2020-09-30 23:26:56 发布 · 565 阅读

0 ·

CC 4.0 BY-SA版权

UVa 专栏收录该内容

46 篇文章

订阅专栏

本文介绍了一种用于预测从完全二叉树根部逐个掉落的K个球最终停靠位置的算法。通过在每个非终端节点设置两个值（真或假）的标志，并根据当前标志值来决定球的移动方向，最终确定球的停靠位置。

Dropping Balls

A number of K balls are dropped one by one from the root of a fully binary tree structure FBT. Each time the ball being dropped first visits a non-terminal node. It then keeps moving down, either follows the path of the left subtree, or follows the path of the right subtree, until it stops at one of the leaf nodes of FBT. To determine a ball's moving direction a flag is set up in every non-terminal node with two values, eitherfalse or true. Initially, all of the flags are false. When visiting a non-terminal node if the flag's current value at this node is false, then the ball will first switch this flag's value, i.e., from thefalse to the true, and then follow the left subtree of this node to keep moving down. Otherwise, it will also switch this flag's value, i.e., from the true to the false, but will follow the right subtree of this node to keep moving down. Furthermore, all nodes of FBT are sequentially numbered, starting at 1 with nodes on depth 1, and then those on depth 2, and so on. Nodes on any depth are numbered from left to right.

For example, Fig. 1 represents a fully binary tree of maximum depth 4 with the node numbers 1, 2, 3, ..., 15. Since all of the flags are initially set to be false, the first ball being dropped will switch flag's values at node 1, node 2, and node 4 before it finally stops at position 8. The second ball being dropped will switch flag's values at node 1, node 3, and node 6, and stop at position 12. Obviously, the third ball being dropped will switch flag's values at node 1, node 2, and node 5 before it stops at position 10.

Fig. 1: An example of FBT with the maximum depth 4 and sequential node numbers.

Now consider a number of test cases where two values will be given for each test. The first value is D, the maximum depth of FBT, and the second one is I, the I^th ball being dropped. You may assume the value of Iwill not exceed the total number of leaf nodes for the given FBT.

Please write a program to determine the stop position P for each test case.

For each test cases the range of two parameters D and I is as below:

$\begin{displaymath}2 \le D \le 20, \mbox{ and } 1 \le I \le 524288.\end{displaymath}$

Input

Contains l +2 lines.


Line 1 		 I the number of test cases 
Line 2 		 
test case #1, two decimal numbers that are separatedby one blank 
... 		 		 
Line k+1 
test case #k 
Line l+1 
test case #l 
Line l+2 -1 		 a constant -1 representing the end of the input file

Output

Contains l lines.


Line 1 		 the stop position P for the test case #1 
... 		 
Line k the stop position P for the test case #k 
... 		 
Line l the stop position P for the test case #l

Sample Input

Sample Output

首先来说一个最经典的二叉树思路，这个代码跑完的结果绝对是正确的，不过需要大量的时间并且占用大量的内存，这玩意就落到下乘了，拿来只是瞅瞅就算了，UVa给的结果是超时。

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;

bool s[ (1<<20) + 8 ];

int main()
{
    int N;
    while(cin >> N && N!=-1)
    {
        while(N--)
        {
            int D,I;
            scanf("%d %d",&D,&I);
            memset(s,0,sizeof(s));
            int k,n= (1<<D) -1;
            for(int i=0;i<I;i++)
            {
                k=1;
                for(;;)
                {
                    s[k]=!s[k];
                    if(s[k]) k*=2;
                    else k*=2,k++;
                    if(k>n) break;
                }
            }
            printf("%d\n",k/2);
        }
    }
    return 0;
}

下面这个代码好，特简单。直接模拟最后一个球球，不过有点难思考。

#include<stdio.h>
int main()
{
    int N;
    while(~scanf("%d",&N)&& N!=-1)
    {
        while(N--)
        {
            int D,I;
            scanf("%d %d",&D,&I);
            int k=1,i;
            for(i=0;i<D-1;i++)
                if(I%2){k*=2;I=(I+1)/2;}
                else {k*=2,k++;I/=2;}
            printf("%d\n",k);
        }
    }
    return 0;
}