《算法竞赛入门经典》6-6小球下落——二叉树的编号

6.3.1 二叉树的编号

6-6 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, either false 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 the false 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 I will 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:
Input
Contains l + 2 lines.
2≤D≤20, and1≤I≤524288.
Line 1 l the number of test cases
Line 2 D1 I1 test case #1, two decimal numbers that are separated by one blank …
Line k+1 Dk Ik test case #k
Line l+1 Dl Il test case #l
Line l + 2 -1 a constant ‘-1’ representing the end of the input file
Output
Contains l lines.
Line 1 … Line k … Line l
the stop position P for the test case #1 the stop position P for the test case #k the stop position P for the test case #l
Sample Input
5 42 34 10 1 22 8 128 -1
Sample Output
12 7 512 3 255

对于一个节点k,其左子节点、右子节点的编号分别是2k和2k+1.

#include 
#include 
#include 
#include 
#include //双向链表
#include //栈
#include //队列
using namespace std;
const int maxd=20;
int s[1<n)break;//已经落“出界”了
            }
        }
        printf("%d\n",k/2);//“出界”之前叶子的编号
    }
    
    return 0;
}

代码缺点：运算量太大。由于I可以高达2的D次方-1个
每个小球都会落在根节点上，因此前2个球必然有一个在左子树上，一个在右子树上。一般的，只需要看小球编号的奇偶性，就能最终知道它到哪颗子树上。对于那些落入跟节点左子树上的小球来说，只需知道该小球是第几个落在根的左子树里的。
当I是奇数时，它是往左走的第（I+1）/2个小球；当I是偶数时，它是往右走的第I/2个小球。这样，可以直接模拟最后一个小球的路线：

int k=1;
for (int i=0; i