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UVA 712(二叉树模拟)
Description
S-Trees
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)
with . 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.
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, , 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 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; }