UVa 712/POJ 1105/ZOJ 1150 S-Trees(用数组模拟二叉树)

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712 - S-Trees

Time limit: 3.000 seconds

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=104&page=show_problem&problem=653

http://poj.org/problem?id=1105

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=1150

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 theroot 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 x_i from the variable set X_n. 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 x_i1 corresponding to the root, a unique variable x_i2 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 x_i, 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.

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, (  x ₁  = 1,  x ₂  = 1  x ₃  = 0) would be a valid VVA form  = 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  x i ₁   x i ₂  ... x i _n . (There will be exactly  n  different space-separated strings). So, for  n  = 3 and the variable ordering  x ₃ ,  x ₁ ,  x ₂ , 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 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   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:
0011S-Tree #2:
0011

英语阅读题=v=:

注意：Depending on whether the value of the variable is 1 or 0, you go its right or left child

变量为1，去右孩子；变量为0，去左孩子。

思路：

建一个二叉树或用数组也行。

(由于数据范围较小:n<=7，这里用数组要好些)

完整代码：（太慢了，没到0ms）

/*UVaOJ: 0.016s*/
/*POJ: 16ms,144KB*/
/*ZOJ: 0ms,180KB*/#include<cstdio>
#include<cstring>
using namespace std;
int serial[7];
char ch[1<<7],temp[7];int main()
{int cas = 1;while (true){int n, t, size;scanf("%d", &n);if (!n)break;getchar();//别忘了读去'\n'//读取x3 x1 x2这样的getchar();for (int i = 0; i < n - 1; i++){scanf("%d", &serial[i]);--serial[i];//这里要减一getchar();getchar();}scanf("%d", &serial[n - 1]);--serial[n - 1];getchar();//别忘了读去'\n'//读取00010011size = 1 << n;for (int i = 0; i < size; i++)ch[i] = getchar();printf("S-Tree #%d:\n", cas++);scanf("%d", &t);getchar();//别忘了读去'\n'while (t--){for (int i = 0; i < n; i++)temp[i] = getchar();getchar();//别忘了读去'\n'int l = 0, r = size, m = size >> 1;for (int i = 0; i < n; i++){if (temp[serial[i]] == '0')r = m;elsel = m;m = (l + r) >> 1;}printf("%c", ch[l]);}printf("\n\n");}return 0;
}

Source

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