UVA 712 (13.08.23)

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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 theroot node and consists of n+1 nodes. Each of the S-tree's nodes has adepth, which is the amount of nodes between itself and the root (so the root has depth 0). The nodes with depth less thann are called non-terminal nodes. All non-terminal nodes have two children: theright 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 variablex_i₂ corresponding to the nodes with depth 1, andso on. The sequence of the variables $x_{i_1}, x_{i_2}, \dots, x_{i_n}$ is called thevariable 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 $x_1, x_2, \dots, x_n$ ,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 variablex_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 isx₃, x₁,x₂.

The values of the variables $x_1, x_2, \dots, x_n$ ,are given as aVariable 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 integern, $1 \le n \le 7$ ,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 exactlyn 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 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 bym 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 ofx₁, 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 givenm VVAs, where f is thefunction 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

题意及做法:

变相的遍历题, 先输入层数n, 然后n个次序(后来证明, 这个输入是废物)

接着就是最后的一层叶子结点 (叶子结点的个数, 其实就是2的n次方)

然后输入一个数m, 代表m次遍历

随后的m行是遍历指令, 0代表往左, 1代表往右~

把m次遍历得到的结果保存下来一次性输出!

要点:

用数组存树的话, 设结点编号是k, 左儿子编号则为2*k, 右儿子编号为2*k+1;

我是一开始就把k设为1, 求出最后叶子结点的编号, 注意了, 此时的编号k不是在数组中的下标, 要减去上面结点数才行

AC代码:

#include<stdio.h>#include<string.h>char order[10][5];char node[512];char ans[10000];int main() {    int n;    int cas = 0;    while(scanf("%d", &n) != EOF && n) {        int i, j;        int presum = 1;        for(i = 0; i < n; i++) {            scanf("%s", order[i]);            presum = presum * 2;        }        getchar();        gets(node);        int m;        scanf("%d", &m);        getchar();        int pos = 0;        char tmp[10];        for(i = 0; i < m; i++) {            gets(tmp);            int len = strlen(tmp);            int k = 1;            for(j = 0; j < len; j++) {                if(tmp[j] == '0')                    k = 2*k;                else                    k = 2*k+1;            }            k = k - presum;            ans[pos++] = node[k];        }        ans[pos] = '\0';        printf("S-Tree #%d:\n", ++cas);        puts(ans);        printf("\n");    }    return 0;}