POJ 3295 Tautology （构造题）

字母：K, A, N, C, E 表示逻辑运算

字母：p, q, r, s, t 表示逻辑变量 0 或 1

给一个字符串代表逻辑表达式，如果是永真式输出tautology 否则输出not

枚举每个逻辑变量的值，5个变量，共2^5种情况，对于每种情况都为真则为永真式。

代码：

/***************************************
Problem: 3295		User:
Memory: 688K		Time: 0MS
Language: G++		Result: Accepted
***************************************/
#include <iostream>
#include <cstring>
#include <cstdio>
#include <stack>

using namespace std;

int p, q, r, s, t; // variables 0 or 1

int st[32][5] = {0,0,0,0,0, 0,0,0,0,1, 0,0,0,1,0, 0,0,0,1,1, 0,0,1,0,0, 0,0,1,0,1, 0,0,1,1,0,
				0,0,1,1,1, 0,1,0,0,0, 0,1,0,0,1, 0,1,0,1,0, 0,1,0,1,1, 0,1,1,0,0, 0,1,1,0,1,
				0,1,1,1,0, 0,1,1,1,1, 1,0,0,0,0, 1,0,0,0,1, 1,0,0,1,0, 1,0,0,1,1, 1,0,1,0,0,
				1,0,1,0,1, 1,0,1,1,0, 1,0,1,1,1, 1,1,0,0,0, 1,1,0,0,1, 1,1,0,1,0, 1,1,0,1,1,
				1,1,1,0,0, 1,1,1,0,1, 1,1,1,1,0, 1,1,1,1,1};
char exp[105];

int get_value(char ch)
{
	switch(ch) {
		case 'p': return p;
		case 'q': return q;
		case 'r': return r;
		case 's': return s;
		case 't': return t;
		case 'N': return -1;
		default: return -2;
	}
}

int WFF(char ch, int a, int b)
{
	if (ch == 'K') return a && b;
	if (ch == 'A') return a || b;
	if (ch == 'C') return (!b) || a;
	if (ch == 'E') return a == b;
}

bool solve()
{
	int i, a, b;
	int len = strlen(exp);
	stack<int> mystack;
	for (i = len - 1; i >= 0; --i) {
		if (get_value(exp[i]) >= 0) {
			a = get_value(exp[i]);
			mystack.push(a);
		} else if (get_value(exp[i]) == -1) {
			a = mystack.top(); mystack.pop();
			a = !a;
			mystack.push(a);
		} else {
			a = mystack.top(); mystack.pop();
			b = mystack.top(); mystack.pop();
			a = WFF(exp[i], b, a);
			mystack.push(a);
		}
	}
	return mystack.top();
}

int main()
{
	int i;
	while (scanf("%s", exp) != EOF) {
		if (exp[0] == '0') break;
		for (i = 0; i < 32; ++i) {
			p = st[i][0]; q = st[i][1]; r = st[i][2];
			s = st[i][3]; t = st[i][4];
			if (!solve()) break;
		}
		if (i == 32) puts("tautology");
		else puts("not");
	}
    return 0;
}