UVa 712/POJ 1105/ZOJ 1150 S-Trees(用数组模拟二叉树)

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712 - S-Trees

Time limit: 3.000 seconds

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=104&page=show_problem&problem=653

http://poj.org/problem?id=1105

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=1150

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 theroot 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, ( x ₁ = 1, x ₂ = 1 x ₃ = 0) would be a valid VVA form = 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 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 $f(x_1, x_2, \dots, x_n)$ 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

英语阅读题=v=:

注意：Depending on whether the value of the variable is 1 or 0, you go its right or left child

变量为1，去右孩子；变量为0，去左孩子。

思路：

建一个二叉树或用数组也行。

(由于数据范围较小:n<=7，这里用数组要好些)

完整代码：（太慢了，没到0ms）

/*UVaOJ: 0.016s*/
/*POJ: 16ms,144KB*/
/*ZOJ: 0ms,180KB*/

#include<cstdio>
#include<cstring>
using namespace std;
int serial[7];
char ch[1<<7],temp[7];

int main()
{
	int cas = 1;
	while (true)
	{
		int n, t, size;
		scanf("%d", &n);
		if (!n)
			break;
		getchar();//别忘了读去'\n'

		//读取x3 x1 x2这样的
		getchar();
		for (int i = 0; i < n - 1; i++)
		{
			scanf("%d", &serial[i]);
			--serial[i];//这里要减一
			getchar();
			getchar();
		}
		scanf("%d", &serial[n - 1]);
		--serial[n - 1];
		getchar();//别忘了读去'\n'

		//读取00010011
		size = 1 << n;
		for (int i = 0; i < size; i++)
			ch[i] = getchar();

		printf("S-Tree #%d:\n", cas++);
		scanf("%d", &t);
		getchar();//别忘了读去'\n'
		while (t--)
		{
			for (int i = 0; i < n; i++)
				temp[i] = getchar();
			getchar();//别忘了读去'\n'
			int l = 0, r = size, m = size >> 1;
			for (int i = 0; i < n; i++)
			{
				if (temp[serial[i]] == '0')
					r = m;
				else
					l = m;
				m = (l + r) >> 1;
			}
			printf("%c", ch[l]);
		}
		printf("\n\n");
	}
	return 0;
}

Source

Mid-Central European Regional Contest 1999

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