zoj S-Trees 1150 （二叉树模拟）
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
#include<stdio.h>
#include<string.h>
#include<math.h>
char str[10][2];//存放树的分支 如（x1 x2 x3）
char a[200];//存放树的末端的数
char b[10];//存放找点的顺序
char c[200];//存放最终找到的一串数
int main(){
int n,m,i,j,l,k,s=0;
while(scanf("%d",&n),n)
{
k=0;
int t=pow(2.0,n);
for(i=0;i<n;i++)
scanf("%s",str[i]);
scanf("%s",a+t);//技巧，从a[t]开始存放
scanf("%d",&m);
for(i=0;i<m;i++)
{
l=1;
scanf("%s",b);
for(j=0;j<n;j++)
{
if(b[j]=='0') //重点：：找最终对应的数
l*=2;
else
l=l*2+1;
}
c[k++]=a[l];
}
//c[k]='\0';//直接输出完毕就结束，对c[k]进行处理
//printf("S-Tree #%d:\n",++s);
//printf("%s\n\n",c);
printf("S-Tree #%d:\n",++s);//这两种输出方式都行
for(i=0;i<k;i++)
printf("%c",c[i]);
printf("\n\n");
}
return 0;
}