poj 3295 Tautology(栈)

题目链接：http://poj.org/problem?id=3295

思路分析：判断逻辑表达式是否为永真式问题。根据该表达式的特点，逻辑词在逻辑变量前，类似于后缀表达式求值问题。

算法中使用两个栈，从表达式的后边开始处理表达式中每个字符；若为逻辑变量，使其入栈SR，否则从栈SR中弹出两个逻辑变量，

进行运算后的结果再入栈SR；直到处理完表达式所有的字符。(PS：使用栈可以很好的处理广义表类似的序列)

 

代码如下：

#include <iostream>
#include <stack>
using namespace std;

const int MAX_N = 1000 + 10;
char A[MAX_N];

int p, q, r, s, t, ans;

int Match(char x)
{
    int value = 0;

    switch(x)
    {
        case '0':value = 0; break;
        case '1':value = 1; break;
        case 'p':value = p; break;
        case 'q':value = q; break;
        case 'r':value = r; break;
        case 's':value = s; break;
        case 't':value = t; break;
        default: value = p;
    }
    return value;
}

int Calc(char a, char b, char T)
{
    int x, y, ans;

    x = Match(a);
    y = Match(b);
    switch(T)
    {
        case 'K':ans = x && y;  break;
        case 'A':ans = x||y;    break;
        case 'N':ans = !x;      break;
        case 'C':ans = (!x)||y; break;
        case 'E':ans = (x == y);break;
    }
    return ans;
}

int WWF( )
{
    int Rusult = 0;
    int Len = strlen(A);
    stack<char>SL, SR;

    for (p = 0; p <= 1; ++ p)
        for(q = 0; q <= 1; ++ q)
            for(r = 0; r <= 1; ++ r)
                for(s = 0; s <=1; ++ s)
                    for(t = 0; t <= 1; ++ t){

        for (int i = 0; i < Len; ++ i)
                SL.push(A[i]);

            while (!SL.empty())
            {
                char x, y, T;
                int flag = 0;

                T = SL.top();SL.pop();
                switch(T)
                {
                    case 'p':
                    case 'q':
                    case 'r':
                    case 's':
                    case 't': SR.push(T);flag = 1;break;
                }

                if (flag == 1)
                    continue;
                else
                if (T == 'N')
                {
                    x = SR.top();SR.pop();
                    ans = Calc(x, x, T);
                }
                else
                {
                    x = SR.top();SR.pop();
                    y = SR.top();SR.pop();
                    ans = Calc(x, y, T);
                }
                SR.push(ans + '0');
            }
            if (ans == 0)
                return 0;
        }
    return 1;
}

int main()
{
    int flag = 0;

    while (scanf( "%s", A ) != EOF)
    {
        if (A[0] == '0')
            break;
        flag = WWF();
        if ( flag == 0 )
            printf( "not\n" );
        else
            printf( "tautology\n" );
    }
    return 0;
}