uva 712 S-Trees（利用二叉树的特点）

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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 adepth, 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 callednon-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 x_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 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 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 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.

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

题目大意：给一个序列集合{x1, x2, x3, ....,xn}, 集合中不是0就是1 ，然后有一个n层的树，集合的序列表示树的填充方向，（对应后面给出的指令）从跟结点出发，如果那个结点是0，就往左儿子方向走，如果是1就往右儿子方向走。最后落在最后一层的叶子结点上，输出这个数字

解题思路：不用建树，碰到0就*2，碰到1就*2+1.

#include<stdio.h>#include<string.h>#include<math.h>#define N 1000#define M 10int main(){int t = 1, n, m;char str[N], tem[M], q[M];int num[M];while (scanf("%d", &n), n){// Init.memset(str, 0, sizeof(str));memset(tem, 0, sizeof(tem));memset(q, 0, sizeof(q));memset(num, 0, sizeof(num));// Read.for (int i = 0; i < n; i++){scanf("%s", tem);num[i] = tem[1] - '0';}scanf("%s", str);scanf("%d", &m);// Handle.int f = pow(2, n);for (int i = 0; i < m; i++){scanf("%s", tem);int k = 1;for (int j = 0; j < n; j++){if (tem[num[j] - 1] == '1')k = k * 2 + 1;elsek = k * 2;}q[i] = str[k - f];}q[m] = '\0';printf("S-Tree #%d:\n%s\n\n", t++, q);}return 0;}