我有一个合乎逻辑的EX pression,我想评价。 这位前pression可以嵌套,由T(真)或F(假)和括号。 括号(意为逻辑或。 每个人身边两个方面TF(或每个人的身边其他两个组合),应进行AND(逻辑与)。
I have a logical expression that I would like to evaluate. The expression can be nested and consists of T (True) or F (False) and parenthesis. The parenthesis "(" means "logical OR". Two terms TF beside each others (or any other two combinations beside each others), should be ANDED (Logical AND).
例如,前pression:
For example, the expression:
((TFT)T) = true
我需要一种算法来解决此问题。我认为首先将前pression以析取或合取范式,然后我可以很容易地评估EX pression。但是,我无法找到一个算法,恢复正常前pression。有什么建议么?谢谢你。
I need an algorithm for solving this problem. I thought of converting the expression first to disjunctive or conjunctive normal form and then I can easily evaluate the expression. However, I couldn't find an algorithm that normalizes the expression. Any suggestions? Thank you.
问题描述可以在这里找到: https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=2&category=378&page=show_problem&problem=2967
The problem statement can be found here: https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=2&category=378&page=show_problem&problem=2967
编辑:我误解了问题的一部分。在给定的逻辑EX pression,与/或运营商相间每个括号(如果我们要重新present前pression一棵树,然后和/或运营商通过在该子树的深度水平。不过,最初给这么树木在最深层次是和树。我的任务是通过构造树可能评估给定的前pression。 低于此谢谢你的答案澄清问题的正确的要求。
I misunderstood part of the problem. In the given logical expression, the AND/OR operators alternate with every parenthesis "(". If we are to represent the expression by a tree, then the AND/OR operators depend on the the sub-tree's depth-level. However, it's initially given that the trees at the deepest level are AND-trees. My task is to evaluate the given expression possibly by constructing the tree. Thanks for the answers below which clarified the correct requirement of the problem.
推荐答案
我使用的是不同的技术提到的那些解决了这个问题。而我得到了它接受了在线系统的法官。
I solved this problem using a different technique than the ones mentioned. And I got it Accepted by the online system judge.
搞清楚运营商在树的第一层后(感谢@Asiri拉斯纳亚克他的想法),我递归构建EX pression树。在施工过程中,我扫描的字符串。如果字符'(',然后我创建了当前运营商值的节点,并把它添加到树中。然后,我交替运营商去进行更深层次的递归级别,如果字符是'T',然后创建与值的节点真,它如果添加到树,并继续扫描。如果字符'F',然后创建,其值为假的一个节点,将其添加到树,并继续扫描。最后,字符是')',然后我回到递归的上一层。
After figuring out the operator at the first level of the tree (Thanks to @Asiri Rathnayake for his idea), I recursively construct the expression tree. During the construction, I scan the string. If the character is '(', then I create a node with the current operator value and add it to the tree. Then, I alternate the operator and go for a deeper recursion level. If the character is 'T', then I create a node with value "True", add it to the tree and continue scanning. If the character is 'F', then I create a node with the value "False", add it to the tree and continue scanning. Finally, if the character is ')', then I return to one level up of the recursion.
最后,我将前pression树完成。现在,所有我需要做的是利用基本的递归函数树一个简单的评测。
At the end, I will have the expression tree completed. Now, all I need to do is a simple evaluation for the tree using basic recursive function.
下面是我的C ++ code:
Below is my C++ code:
#include<iostream>
#include<string>
#include<vector>
#include<algorithm>
using namespace std;
struct Node {
char value;
vector<Node*> children;
};
void ConstructTree (int &index, string X, Node *&node, int op)
{
for(; index<X.size(); index++)
{
if(X[index]=='T')
{
Node *C= new Node;
C->value='T';
node->children.push_back(C);
}
else if(X[index]=='F')
{
Node* C= new Node;
C->value='F';
node->children.push_back(C);
}
else if(X[index]=='(')
{
if(op==0)
{
Node* C= new Node;
C->value='O';
node->children.push_back(C);
}
else
{
Node* C= new Node;
C->value='A';
node->children.push_back(C);
}
index++;
ConstructTree(index,X,node->children[node->children.size()-1],1-op);
}
else
return;
}
}
bool evaluateTree(Node* node)
{
if(node->value=='T')
return true;
else if(node->value=='F')
return false;
else if(node->value=='O')
{
for(int i=0; i<node->children.size(); i++)
if(evaluateTree(node->children[i])==true)
return true;
return false;
}
else if(node->value=='A')
{
for(int i=0; i<node->children.size(); i++)
if(evaluateTree(node->children[i])==false)
return false;
return true;
}
}
int main()
{
string X;
int testCase=1;
while(cin>>X)
{
if(X=="()")
break;
int index=0;
int op=-1;
int P=0;
int max=0;
for(int i=0; i<X.size(); i++)
{
if(X[i]=='(')
P++;
if(X[i]==')')
P--;
if(P>max)
max=P;
}
if(max%2==0)
op=0; //OR
else
op=1; //AND
Node* root = new Node;
if(op==0)
root->value='O';
else
root->value='A';
index++;
ConstructTree(index,X,root,1-op);
if(evaluateTree(root))
cout<<testCase<<". true"<<endl;
else
cout<<testCase<<". false"<<endl;
testCase++;
}
}
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