POJ 3295 Tautology（构造法）

http://poj.org/problem?id=3295

题意：

判断表达式是否为永真式。

思路：

把每种情况都枚举一下。

 #include<iostream>
 #include<string>
 #include<cstring>
 using namespace std;

 const int MAXN = ;

 int sta[MAXN];
 char str[MAXN];
 int p, q, r, s, t;

 void  judge()
 {
     int top = ;
     int len = strlen(str);
     for (int i = len - ; i >= ; i--)
     {
         if (str[i] == 'p') sta[top++] = p;
         else if (str[i] == 'q') sta[top++] = q;
         else if (str[i] == 'r') sta[top++] = r;
         else if (str[i] == 's') sta[top++] = s;
         else if (str[i] == 't') sta[top++] = t;
         else if (str[i] == 'K')
         {
             int t1 = sta[--top];
             int t2 = sta[--top];
             sta[top++] = (t1&&t2);
         }
         else if (str[i] == 'A')
         {
             int t1 = sta[--top];
             int t2 = sta[--top];
             sta[top++] = (t1 || t2);
         }
         else if (str[i] == 'N')
         {
             int t1 = sta[--top];
             sta[top++] = (!t1);
         }
         else if (str[i] == 'C')
         {
             int t1 = sta[--top];
             int t2 = sta[--top];
             if (t1 ==  && t2 == )sta[top++] = ;
             else sta[top++] = ;
         }
         else if (str[i] == 'E')
         {
             int t1 = sta[--top];
             int t2 = sta[--top];
             if ((t1 ==  && t2 == ) || (t1 ==  && t2 == )) sta[top++] = ;
             else sta[top++] = ;
         }
     }
 }

 bool solve()
 {
     for (p = ; p<; p++)
     for (q = ; q<; q++)
     for (r = ; r<; r++)
     for (s = ; s<; s++)
     for (t = ; t<; t++)
     {
         judge();
         if (sta[] == )return false;
     }
     return true;
 }

 int main()
 {
     //freopen("D:\\txt.txt", "r", stdin);
     while (gets(str) && str[] != '')
     {
         if (solve())  cout << "tautology" << endl;
         else          cout << "not" << endl;
     }
     return ;
 }