1105-S-Trees


题目大意:

给一棵完整的树:

给出每一层的节点号(xi,同一层的节点共用一个号,从第一层到第n层不一定是从x1到xn,是打乱的)

给出所有叶子节点的值(如n=3,叶子节点有8个)

给出路径v1v2v3...Vn:0走左子树,1走右子树,但是路径是严格分配给x1x2x3...xn的,即如果第一层节点即根节点不是x1而是x3,走的第一步要根据路径的v3来确定左子树还是右子树

输出最后到达的叶子节点的值

思想:

到达每一个叶子节点的路径是唯一的

s=s*2+(vvn[i][j]-‘0‘);

数据结构的知识

父节点是i,左孩子是i*2,右孩子是2*i+1;

在这题中只存储叶子节点即可





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Language:Default

S-Trees


Time Limit: 1000MS


Memory Limit: 10000K


Total Submissions: 1439


Accepted: 773

Description

A Strange Tree  (S-tree) over the variable set Xn = {x1,x2,...,xn} is a binary tree  representing a Boolean function f:{0,1}->{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 xi1,xi2,...,xin 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 x1,x2,...,xn, then it is quite simple  to find out what f(x1,x2,...,xn) 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 x1 and (x2 or x3) function

On the picture, two S-trees representing the same Boolean  function,f(x1,x2,x3) = x1 and (x2 or x3), 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 x1,x2,...,xn, are given as a Variable Values  Assignment (VVA)

(x1 = b1, x2 = b2,  ..., xn = bn)

with b1,b2,...,bn 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 and (1 or 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(x1,x2,...,xn) as described above.

Input

The input 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  <= n <= 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 xi1 xi2  ...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 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 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(x1,x2,...,xn) 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

Source

Mid-Central European  Regional Contest 1999

[Submit]  [Go Back]   [Status]   [Discuss]

代码:

# include <iostream>

# include <cstdio>

# include <string>

# include <cmath>

using namespace std;

int main()

{

int n,nvvn,i,j,s,m=0;

char ord[30],tnod[300];

cin>>n;

while(n!=0)

{

m++;

int* order=new int[n];

getchar();

gets(ord);

for(i=0;i<n;i++)

{

order[i]=(ord[1+i*3]-‘0‘)-1;

}

cin>>tnod;

cin>>nvvn;

char (*vvn)[10]=new char[nvvn][10];

for(i=0;i<nvvn;i++)

cin>>vvn[i];

cout<<"S-Tree #"<<m<<":"<<endl;

for(i=0;i<nvvn;i++)

{

for(j=0,s=0;j<n;j++)

{

s=s*2+(vvn[i][j]-‘0‘);

}

cout<<tnod[s];

}

cout<<endl;

cout<<endl;

cin>>n;

}

return 0;

}

时间: 2024-11-06 22:49:11

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