uva 712 - S-Trees

最新推荐文章于 2018-02-06 20:43:00 发布

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最新推荐文章于 2018-02-06 20:43:00 发布

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原文链接：http://www.cnblogs.com/jecyhw/p/3903020.html

本文深入解析S-Tree的概念及其在布尔函数表示中的应用，通过实例展示了如何将变量赋值转化为对S-Tree的遍历过程，进而求得布尔函数的输出值。文中详细解释了S-Tree的结构、变量排序、终端节点的值分布，并提供了实现该功能的程序代码。通过具体的输入输出示例，清晰地展示了从输入到输出的完整流程。

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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 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 $x_{i_1}, x_{i_2}, \dots, x_{i_n}$ 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 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 $f(x_1, x_2, \dots, x_n)$ 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

Sample Output

S-Tree #1:
0011

S-Tree #2:
0011

#include <iostream>
#include <stack>
#include <cstring>
#include <cstdio>
#include <string>
#include <algorithm>
#include <queue>
#include <set>
#include <map>
#include <fstream>
#include <stack>
#include <list>
#include <sstream>
#include <cmath>

using namespace std;
/*

*/
#define ms(arr, val) memset(arr, val, sizeof(arr))
#define N 10
#define INF 0x3fffffff
#define vint vector<int>
#define setint set<int>
#define mint map<int, int>
#define lint list<int>
#define sch stack<char>
#define qch queue<char>
#define sint stack<int>
#define qint queue<int>

int stree[N];
char term[1000];
char vva[N];
/*
看懂了题目还是比较简单的。
给一个vva，找对应的叶子节点的值。
vva中0表示往左子树走，1表示在右子树找
例如：
题目中的right tree it is x3, x1, x2；
给的vva是1 1 0，也就是x3=0,x1=1,x2=2
那么找的话，就是先左再右再右就是答案1了
*/
int expo(int a, int n)//a^n
{
    int ans = 1;
    while (n)
    {
        if (n & 1)
        {
            ans *= a;
        }
        a *= a;
        n >>= 1;
    }
    return ans;
}
void bin_search(int i, int j, int p)//二分搜索
{
    if (i >= j)
    {
        putchar(term[j]);
        return;
    }
    if (vva[stree[p]] == '0')
        bin_search(i, (i + j) / 2, p + 1);
    else
        bin_search((i + j) / 2 + 1, j, p + 1);
}
int main()
{
    int n, len, i, m, cases = 1;
    string t;
    while (scanf("%d", &n), n)
    {
        i = 1;
        cin>>t;
        len = t.length() - 1;
        stree[i] = t[len] - '0';
        for (i++; i <= n; i++)
        {
            cin>>t;
            stree[i] = t[len] - '0';
        }
        len = expo(2, n);
        scanf("%s", term + 1);
        scanf("%d", &m);
        printf("S-Tree #%d:\n", cases++);
        while (m--)
        {
            scanf("%s", vva + 1);
            bin_search(1, len, 1);
        }
        printf("\n\n");
    }
    return 0;
}