首页 > 代码库 > UVA 712(二叉树模拟)

UVA 712(二叉树模拟)

2024-07-06 17:26:26   222人阅读

L - S-Trees
Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu
Submit Status
Appoint description: 

Description

Download as PDF


  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 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 $x_{i_1}, x_{i_2}, \dots, x_{i_n}$ 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 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 $x_1 \wedge (x_2 \vee x_3)$

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 ... 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 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 exactlyn 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 

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


Miguel A. Revilla
2000-02-09

就是给出n个变量，变量为1往右走，为0往左走，问到达最底层时的值是多少，数组模拟即可。

#include<stdio.h>
#include<string.h>
char ans[1005],temp[2005];
int n,o[1<<9],num;
char s[1<<9],b[1<<9];
int main()
{
    int cas=1;
    //freopen("in.txt","r",stdin);
    while(~scanf("%d",&n)&&n)
    {
        getchar();
        gets(temp);
        int len=strlen(temp);
        num=0;
        int k=1,ok=0;
        for(int i=0;i<=len;i++)
        {
            if((temp[i]>='0'&&temp[i]<='9'))
            {
                num+=num*10+temp[i]-'0';
                ok=1;
            }
            else if(ok)
            {
                o[k++]=num;
                //printf("%d\n",num);
                ok=num=0;
            }
        }
        int m;
        scanf("%s%d",s+1,&m);
        int t=0;
        while(m--)
        {
            int k=1;
            scanf("%s",b+1);
            for(int i=1;i<=n;i++)
            {
                if(b[o[i]]-'0')k=2*k+1;
                else k=2*k;
            }
            k-=(1<<n)-1;
            ans[t++]=s[k];
        }
        ans[t]='\0';
        printf("S-Tree #%d:\n%s\n\n",cas++,ans);
    }
    return 0;
}


联系
我们