UVA

  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 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_i1corresponding to the root, a unique variable x_i2 corresponding to the nodes with depth 1, and so 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 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 

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 is x₃, x₁, x₂.

The values of the variables , are given as a Variable Values Assignment (VVA) 

with . 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 computes  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, , 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 xi₁xi₂ ... 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  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 

3
x1 x2 x3
00000111
4
000
010
111
110
3
x3 x1 x2
00010011
4
000
010
111
110
0

Sample Output 

S-Tree #1:
0011

S-Tree #2:
0011

又是一道水题

#include <bits/stdc++.h>
using namespace std;

int main()
{
    int kase = 0;
    int n, m;
    while(cin >> n, n != 0)
    {
        string x;
        int change[10];
        int tree_val[256];
        for(int i = 0; i < n; i++){
            cin >> x;
            change[i] = x[1] - '0';
        }
        cin >> x;
        for(size_t i = 0; i < x.length(); i++)
            tree_val[i] = x[i] - '0';

        cin >> m;
        vector<int> res;
        while(m--){
            cin >> x;
            int id = pow(2, n), base = pow(2, n-1);
            for(size_t i = 0; i < n; i++){
                if(x[change[i]-1]=='0')
                    id -= base;
                base /= 2;
            }
            res.push_back(tree_val[id-1]);
        }
        printf("S-Tree #%d:
", ++kase);
        for(int i = 0; i < res.size(); i++)
            cout << res[i];
        cout << endl << endl;
    }
    return 0;
}