标签：uva acm data structure binary trees
题目如下：
| S-Trees |
A Strange Tree (S-tree) over the variable set 
is a binary tree representing a Boolean function
.Each path of the S-tree begins at the
root node and consists of n+1 nodes. Each of the S-tree‘s nodes has a
depth, which is the amount of nodes between itself and the root (so the root has depth 0). The nodes with depth less than
n are called non-terminal nodes. All non-terminal nodes have two children: the
right child and the left child. Each non-terminal node is marked with some variable
xi from the variable set Xn. All non-terminal nodes with the same depth are marked with the same variable, and non-terminal nodes with different depth are marked with different variables. So, there is
a unique variable xi1 corresponding to the root, a unique variable
xi2 corresponding to the nodes with depth 1, andso on. The sequence of the variables
is called the
variable ordering. The nodes having depth n are called terminal nodes. They have no children and are marked with either 0 or 1. Note that the variable ordering and the distribution of 0‘s and 1‘s on terminal nodes are sufficient to
completely describe an S-tree.
As stated earlier, each S-tree represents a Boolean function f. If you have an S-tree and values for the variables
,then it is quite simple to find out what
is: start with the root. Now repeat the following: if the node you are at is labelled with a variablexi,
then depending on whether the value of the variable is 1 or 0, you go its right or left child, respectively. Once you reach a terminal node, its label gives the value of the function.
Figure 1: S-trees for the function 
On the picture, two S-trees representing the same Boolean function, 
,are shown. For the left
tree, the variable ordering is x1, x2,
x3, and for the right tree it isx3, x1,
x2.
The values of the variables ,are given as a
Variable Values Assignment (VVA)
.For instance, (x1 = 1,
x2 = 1 x3 = 0) would be a valid VVA for
n = 3, resulting for the sample function above in the value 
.The corresponding paths are shown
bold in the picture.
Your task is to write a program which takes an S-tree and some VVAs and computesas described above.
,the depth of the S-tree. This is followed by a line describing the variable ordering of the S-tree. The format
of that line is xi1 xi2 ...xin. (There will be exactly
n different space-separated strings).So, for n = 3 and the variable ordering
x3, x1, x2, this line would look as follows:
x3 x1 x2
In the next line the distribution of 0‘s and 1‘s over the terminal nodes is given. There will be exactly 2n characters (each of which can be 0 or 1), followed by the new-line character.The characters are given in the order in which they appear in the S-tree, the first character corresponds to the leftmost terminal node of the S-tree, the last one to its rightmost terminal node.
The next line contains a single integer m, the number of VVAs, followed by m lines describing them. Each of the m lines contains exactly n characters (each of which can be 0 or 1), followed by a new-line character. Regardless of the variable ordering of the S-tree, the first character always describes the value of x1, the second character describes the value of x2, and so on. So, the line
110
corresponds to the VVA (x1 = 1, x2 = 1, x3 = 0).
The input is terminated by a test case starting with n = 0. This test case should not be processed.
for each of the given
m VVAs, where f is thefunction defined by the S-tree.
Output a blank line after each test case.
3 x1 x2 x3 00000111 4 000 010 111 110 3 x3 x1 x2 00010011 4 000 010 111 110 0
S-Tree #1: 0011 S-Tree #2: 0011
只要把题意搞清楚了这道题还是很简单的，按照正常思路先建树再遍历，直接模拟即可。遍历的时候遍历两次，第一次赋值，第二次得出结果，输入的时候我是把X忽略，根据X的下标确定每个节点的位置，然后在遍历赋值的时候就可以根据这个下标给节点赋值。把建树放在循环中使每次的数据互不干扰。
AC的代码如下：
UVA 712 - S-Trees,布布扣,bubuko.com
标签：uva acm data structure binary trees
原文地址：http://blog.csdn.net/u013840081/article/details/27365809
