Uva 712 - S-Trees-白红宇

Uva 712 - S-Trees

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发布时间：2019-06-28

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S-Trees

A Strange Tree (S-tree) over the variable set $X_n = \{x_1, x_2, \dots, x_n\}$ is a binary tree representing a Boolean function $f: \{0, 1\}^n \rightarrow \{ 0, 1\}$ . 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 variablex_i from the variable set X_n. 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 x_i₁ corresponding to the root, a unique variable x_i₂ corresponding to the nodes with depth 1, and so on. The sequence of the variables $x_{i_1}, x_{i_2}, \dots, x_{i_n}$ is called the variable ordering. The nodes having depth nare 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 $x_1, x_2, \dots, x_n$ , then it is quite simple to find out what $f(x_1, x_2, \dots, x_n)$ is: start with the root. Now repeat the following: if the node you are at is labelled with a variable x_i, 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 $x_1 \wedge (x_2 \vee x_3)$

On the picture, two S-trees representing the same Boolean function, $f(x_1, x_2, x_3) = x_1 \wedge (x_2 \vee x_3)$ , are shown. For the left tree, the variable ordering is x₁, x₂, x₃, and for the right tree it is x₃, x₁, x₂.

The values of the variables $x_1, x_2, \dots, x_n$ , are given as a Variable Values Assignment (VVA)

$\begin{displaymath}(x_1 = b_1, x_2 = b_2, \dots, x_n = b_n) \end{displaymath}$

with $b_1, b_2, \dots, b_n \in \{0,1\}$ . 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 $f(1, 1, 0) = 1 \wedge (1 \vee 0) = 1$ . 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 computes $f(x_1, x_2, \dots, x_n)$ as described above.

Input

The input file contains the description of several S-trees with associated VVAs which you have to process. Each description begins with a line containing a single integer n, $1 \le n \le 7$ , 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...xi_n. (There will be exactly n different space-separated strings). So, for n = 3 and the variable orderingx₃, 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 2ⁿcharacters (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.

Output

For each S-tree, output the line ``S-Tree #j:", where j is the number of the S-tree. Then print a line that contains the value of $f(x_1, x_2, \dots, x_n)$ for each of the given m VVAs, where f is the function defined by the S-tree.

Output a blank line after each test case.

Sample Input

3x1 x2 x30000011140000101111103x3 x1 x20001001140000101111100

Sample Output

S-Tree #1:0011S-Tree #2:0011

Miguel A. Revilla

2000-02-09

#include
     
      #include
      
       #include
       
        #include
        
         #define MAXN 300int operation(int i, int *tree, int len){
    //找到叶子节点的值并返回     if(2*i+2 >= len) return tree[i];    else    {        if(tree[i] == 0) return operation(2*i+1, tree, len);        else return operation(2*i+2, tree, len);    }}int main(){    int n, i, j, m, sum, temp, flag, t, times, k, cnt = 0;    int s_Tree[MAXN], order[10], read[10];    char ch;    while(scanf("%d", &n) != EOF && n)    {        getchar();        i = flag = 0;        memset(order, 0, sizeof(order));        while((ch = getchar()) != '\n')        {
    //处理输入，将xn中的数放到数组order中             if(isdigit(ch))            {                order[i] = order[i]*10 + (ch - '0');                flag = 1;            }            else if(flag == 1)            {                i++;                flag = 0;            }        }        memset(s_Tree, 0, sizeof(s_Tree));        sum = (int)pow(2.0, (double)n) - 1; //找到叶子节点的存储位置先         m = (int)pow(2.0, (double)n);//叶子节点的数量         for(i=0; i

解题思路：

简单题，所以还是跟以前一样，建树，不过因为这次是层次遍历的模式，所以用数组存储各节点的值，最后根据父和子节点的关系找到最终的值

转载于:https://www.cnblogs.com/liaoguifa/archive/2013/03/25/2981273.html

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