## 712 – S-TreesTime limit: 3.000 seconds |

S-Trees |

*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, 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.

On the picture, two S-trees representing the same Boolean function,, are shown. For the left tree, the variable ordering is *x*_{1}, *x*_{2}, *x*_{3}, and for the right tree it is *x*_{3}, *x*_{1}, *x*_{2}.

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 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 `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 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 *x*_{1}, the second character describes the value of *x*_{2}, and so on. So, the line

`110`

corresponds to the VVA ( *x*_{1} = 1, *x*_{2} = 1, *x*_{3} = 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

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 |
3 x1 x2 x3 00000111 4 000 010 111 110 3 x3 x1 x2 00010011 4 000 010 111 110 0 |

## Sample Output

1 2 3 4 5 |
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。

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 |
/******************************************************** * 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; } |