Description
| S-Trees |
A Strange Tree (S-tree) over the variable set 
is a binary
tree representing a Boolean function 
. 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 
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 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 
On the picture, two S-trees representing the same Boolean function, 
,
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 , are given as a Variable Values Assignment (VVA)
. 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 
. 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.
,
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.
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.
3 x1 x2 x3 00000111 4 000 010 111 110 3 x3 x1 x2 00010011 4 000 010 111 110 0
S-Tree #1: 0011 S-Tree #2: 0011
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;
}
原文地址：http://blog.csdn.net/u013268685/article/details/27678007
