【UVA 712 S-Tree】& 二叉树

A Strange Tree (S-tree) over the variable set Xn = {x1, x2, … , xn} is a binary tree representing a
Boolean function f : {0, 1}
n → {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 function x1 ∧ (x2 ∨ x3)
On the picture, two S-trees representing the same Boolean function, f(x1, x2, x3) = x1 ∧ (x2 ∨ 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 ∈ {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 ∧ (1 ∨ 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 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, 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.

题意: 给出 N 个 xi 的位置，从上往下，以及这棵树的叶子节点，然后 m 次查询，每次查询输出对于位置的叶子节点，每次查询 1 代表左，0 代表右

思路 ： 按给出的 xi 的顺序赋值依次为 2 ^ (n - i),对于每次查询，直接算出权值输出对应位置的字符

AC代码：

#include<cstdio>
#include<cmath>
#include<deque>
#include<queue>
#include<vector>
#include<cstring>
#include<algorithm>
using namespace std;
const int MAX = 1e5 + 10;
const int INF = 1e9 + 7;
typedef long long LL;
char s[MAX],c[110],sl[MAX];
int a[MAX],b[MAX],o[MAX];
int main()
{
    int n,ncase = 0,ok = 0;
    while(~scanf("%d",&n),n){
        for(int i = 1; i <= n; i++)
            scanf("%s",c),a[i] = c[1] - '0';
        LL p = 1;
        for(int i = n; i >= 1; i--)
            o[a[i]] = p,p *= 2;
        scanf("%s",s);
        int m;
        scanf("%d",&m);
        for(int i = 1; i <= m; i++){
            scanf("%s",sl);
            int ans = 0;
            for(int j = 0; j < n; j++)
                if(sl[j] == '1')
                    ans += o[a[j + 1]];
            b[i] = ans;
        }
        printf("S-Tree #%d:\n",++ncase);
        for(int i = 1; i <= m; i++)
            printf("%c",s[b[i]]);
        puts("\n");
    }
    return 0;
}