UVAOJ712 | 哈喽哈咯

Written by razrlele 11:26 February 3, 2015

712 – S-Trees

Time limit: 3.000 seconds

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 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

x1 x2 x3

00000111

000

010

111

110

x3 x1 x2

00010011

000

010

111

110

Sample Output

S-Tree #1:

0011

S-Tree #2:

0011

Miguel A. Revilla
2000-02-09

题目的意思是在深度为n的树中，从root（深度为0）到深度为n-1的节点分别是x1,x2,x3,x4…xn中的一个（互相不重复），然后输入数据的时候首先会给出从root到深度为n-1的节点的xi序列，然后从左往右给出2^n个叶子的值（0或1），最后再依次给出x1,x2,x3的值，其中1代表往右，0代表往左，然后求出每一条路径对应的叶子值即可。

这个题目不需要涉及到建树的操作，从左往右每一个叶子所对应的路径分别是0… 0…1 0…10 0…11，即分别是0到2^n-1的二进制数，因此直接把路径转换成十进制数就可以直接知道叶子的值了。

/********************************************************

* File Name: uvaoj712.cpp

* Author: razrLeLe

* Mail: razrlele@outlook.com

* Homepage: https://yueyu.io

* Created Time: Mon 02 Feb 2015 03:31:54 PM CST

******************************************************/

#include "vector"

#include "set"

#include "deque"

#include "queue"

#include "algorithm"

#include "functional"

#include "iostream"

#include "cstdio"

#include "cmath"

#include "cstdlib"

#include "string"

#include "cstring"

#include "string.h"

#include "map"

#include "cctype"

#include "list"

#include "stack"

using namespace std;

#define INF 0x3f3f3f3f

#define LOCAL

int main()

{

#ifdef LOCAL

freopen("/home/razrlele/build/data.txt", "r", stdin);

#endif

string terminal_nodes;

string in;

string ans;

int depth;

int q;

int x[7];

int *p[7];

int place;

int casenum = 0;

while(cin >> depth && depth)

{

ans = "";

for(int i = 0; i < depth; i++)

{

cin >> in;

p[i] = &x[in[1]-'0'-1];

}

cin >> terminal_nodes;

cin >> q;

for(int i = 0; i < q; i++)

{

cin >> in;

for(int j = 0; j < depth; j++)

{

x[j] = in[j]-'0';

}

place = 0;

for(int j = 0; j < depth; j++)

{

place += pow(2, depth-1-j) * (*p[j]);

}

ans = ans + terminal_nodes[place];

}

printf("S-Tree #%d:\n", ++casenum);

cout << ans << endl << endl;

}

return 0;

}