POJ 3295

问题概述

*输入由 p、q、r、s、t、K、A、N、C、E 共 10 个字母组成的逻辑表达式，
*其中 p、q、r、s、t 的值为 1（true）或 0（false），即逻辑变量；
*K、A、N、C、E 为逻辑运算符，
*K --> and: x && y
*A --> or: x || y
*N --> not : ! x
*C --> implies : (!x)||y
*E --> equals : x==y
*问这个逻辑表达式是否为永真式。
*PS:输入格式保证是合法的

代码如下

//3295
//Tautology

#include<iostream>
#include<stack>
#include<bitset>


using namespace std;

const string Variables = "pqrst";
const string Operators = "KANCE";


//掩码
const short Mask_p = 1;
const short Mask_q = 2;
const short Mask_r = 4;
const short Mask_s = 8;
const short Mask_t = 16;

bool calculate( stack<bool> &operands, char operation )
{
	bool ans, op1, op2;

	op1 = operands.top();
	operands.pop();

	//单目运算
	if( operation == 'N' ){
		ans = !op1;
		//cout << op1 << operation << endl;	//test
	}
	else{	//双目运算
		op2 = operands.top();
		operands.pop();

		switch( operation ) {
			case 'K':
				ans = op1 && op2;
				break;
			case 'A':
				ans = op1 || op2;
				break;
			case 'C':
				ans = (!op2) || op1;
				break;
			case 'E':
				ans = op1 == op2;
				break;
		}

		//cout << op1 << operation << op2 << endl;	//test
	}

	operands.push(ans);

	return ans;
}


bool changeToBool( short vars, char var )
{
	bool ans;

	switch( var ){
		case 'p':
			ans = vars & Mask_p;
			break;
		case 'q':
			ans = vars & Mask_q;
			break;
		case 'r':
			ans = vars & Mask_r;
			break;
		case 's':
			ans = vars & Mask_s;
			break;
		case 't':
			ans = vars & Mask_t;
			break;
	}

	return ans;
}


int main()
{
	const short range = 0x1f;
	short vars = 0;
	stack<bool> operands;

	string formula;


	while( cin >> formula, formula != "0" ){

		for( vars = 0;  vars <= range;  vars++ ){
			//cout<< vars << " = " << bitset<sizeof(short)*8>(vars)  << endl;	//test
			for( string::reverse_iterator c = formula.rbegin();  c != formula.rend();  c++ ){	//注意c的类型
				if( Variables.find(*c) != string::npos ){
					operands.push( changeToBool(vars, *c) );
				}else{
					calculate( operands, *c );	
				}
			}

			if( operands.top() == false ){
				cout << "not" << endl;
				break;
			}

		}

		if( operands.top() == true )
			cout << "tautology" << endl;

		while( !operands.empty() )	//清空栈
			operands.pop();
	
	}

	return 0;
}

注意

• 使用了string::reverse_iterator
• 首先审题，注意题目是只输入单个数据就结束还是循环输入
• 分别处理好程序正确与错误的结束方式
• 不好搞混变量，分析最有可能出错的地方

---

这题的考点我觉得主要是位操作和前缀表达式的运行。而对于前缀表达式运算，上面的方法是自底而上的思想，还可以用递归的方式进行自顶向下的运算。

//3295
//Tautology
//递归求解

#include<iostream>
#include<stack>
#include<bitset>


using namespace std;

const string Variables = "pqrst";
const string Operators = "KANCE";


//掩码
const short Mask_p = 1;
const short Mask_q = 2;
const short Mask_r = 4;
const short Mask_s = 8;
const short Mask_t = 16;


int cnt = -1;	
bool step( string formula, short vars )
{
	bool ans;
	cnt++;	//必须放在前面

	switch( formula[cnt] ){
		case 'p':
			ans = vars & Mask_p;
			break;
		case 'q':
			ans = vars & Mask_q;
			break;
		case 'r':
			ans = vars & Mask_r;
			break;
		case 's':
			ans = vars & Mask_s;
			break;
		case 't':
			ans = vars & Mask_t;
			break;
		case 'N':
			ans = !step(formula, vars);
			break;
		case 'K':
			//只能是&而不能是&&，因为&&有短路特性，可能会跳过第二个step
			ans = step(formula, vars) & step(formula, vars);
			break;
		case 'A':
			ans = step(formula, vars) | step(formula, vars);
			break;
		case 'C':
			ans = ( !step(formula, vars) )  |  step(formula, vars);
			break;
		case 'E':
			ans = step(formula, vars) == step(formula, vars);
			break;
			
	}

	return ans;
}

int main()
{
	const short range = 0x1f;
	short vars = 0;
	bool ans;

	string formula;


	while( cin >> formula, formula != "0" ){

		//cout << formula << endl;	//test
		for( vars = 0;  vars <= range;  vars++ ){

			cnt = -1;
			ans = step( formula, vars );

			if( ans == false ){
				cout << "not" << endl;
				break;
			}
		}

		if( ans == true )
			cout << "tautology" << endl;

	}


	return 0;
}