UVA 712 S-Trees
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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 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 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
3x1 x2 x30000011140000101111103x3 x1 x20001001140000101111100
Sample Output
S-Tree #1:0011S-Tree #2:0011题意:输入n代表树有n层 接着输入xi代表每层的名字。。然后输入一串代表叶子节点。
然后输入一个k,接下来输入k个串,代表遍历顺序。。0往左,1往右。
这题直接用数组模拟就可以了。。存下叶子节点。
然后根据遍历顺序。遇到1就在叶子节点往右移动 2 ^(总层数 - 当前层数) 个位置。。
然后逐个输出就可以了。。
#include <stdio.h>#include <string.h>#include <math.h>int n;int m;int i;int x[10];char dd[10];int d[10];char ddi[205];int di[205];int main(){int tt = 1;while (scanf("%d", &n) != EOF && n){getchar();for (i = 1; i <= n ; i++){int sbb;scanf("%*c%d%*c", &sbb);x[sbb] = i;}gets(ddi);int lenddi = strlen(ddi);for (i = 1; i <= lenddi; i ++)di[i] = ddi[i - 1] - '0';printf("S-Tree #%d:\n", tt ++);scanf("%d", &m);getchar();while (m --){gets(dd);memset(d, 0, sizeof(0));int lendd = strlen(dd);for (i = 1; i <= lendd; i ++){d[x[i]] = dd[i - 1] - '0';}int sb = 1;for (i = n; i >= 1; i --){if (d[i] == 1){sb += int(pow(2, n - i) + 0.5);}}printf("%d", di[sb]);}printf("\n\n");}return 0;}
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