S-Trees

A Strange Tree (S-tree) over the variable set Xn = {x1, x2, . . . , xn} is a binary tree representing a Boolean function f : {0, 1} n → {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 variable xi from the variable set Xn. 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 xi1 corresponding to the root, a unique variable xi2 corresponding to the nodes with depth 1, and so on. The sequence of the variables xi1 , xi2 , . . ., xin 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 x1, x2, . . ., xn, then it is quite simple to find out what f(x1, x2, . . . , xn) is: start with the root. Now repeat the following: if the node you are at is labelled with a variable xi , 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 x1 ∧ (x2 ∨ x3) On the picture, two S-trees representing the same Boolean function, f(x1, x2, x3) = x1 ∧ (x2 ∨ x3), 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 x1, x2, . . ., xn, are given as a Variable Values Assignment (VVA) (x1 = b1, x2 = b2, . . . , xn = bn) with b1, b2, . . . , bn ∈ {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 ∧ (1 ∨ 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(x1, x2, . . . , xn) 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 ≤ n ≤ 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 . . . xin . (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 2 n 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(x1, x2, . . . , xn) 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

题意：给出n层节点的顺序，然后给出叶节点的值，有q次查询，每次查询给出路径，0代表向左走，1代表向右走，问走完后到达的叶子节点的值sh是多少？

思路：这个题做的时候开一个数组，数组中每个元素对应一个叶子节点的值，明白向左走是2*k，向右走是2*k+1，然后找到得到的下标对应的元素即可。代码如下：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int main()
{
    int n,i,j,l,t,k,m=1;
    int num[2050],tt[2050];
    char s[3],str[2050],ss[10];
    while(~scanf("%d",&n)&&n)
    {
        for(i=1; i<=n; i++)
            scanf("%s",s);
        scanf("%s",str);
        l=strlen(str);
        for(i=l; i<2*l; i++)
            num[i]=str[i-l]-'0';
        int h=0;
        scanf("%d",&t);
        for(i=0; i<t; i++)
        {
            scanf("%s",ss);
            k=1;
            for(j=0; j<n; j++)
            {
                if(ss[j]=='0')
                    k=k*2;
                else
                    k=k*2+1;
            }
            tt[h++]=num[k];
        }
        printf("S-Tree #%d:\n",m++);
        for(i=0;i<h;i++)
            printf("%d",tt[i]);
            printf("\n\n");
        }
    return 0;
}