UVA 712 (13.08.23)

 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_i1 corresponding to the root, a unique variable x_i2 corresponding to the nodes with depth 1, andso 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 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

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 isx₃, x₁, x₂.

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

with .For instance, ( x ₁ = 1, x ₂ = 1 x ₃ = 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 computesas 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 x i ₁ x i ₂ ... x i _n. (There will be exactly n different space-separated strings).So, for n = 3 and the variable ordering x ₃, x ₁, x ₂, 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 thefunction 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] = '';
        printf("S-Tree #%d:
", ++cas);
        puts(ans);
        printf("
");
    }
    return 0;
}