poj 3295 Tautology (模拟栈操作+状压）

WFF 'N PROOF is a logic game played with dice. Each die has six faces representing some subset of the possible symbols K, A, N, C, E, p, q, r, s, t. A Well-formed formula (WFF) is any string of these symbols obeying the following rules:

• p, q, r, s, and t are WFFs
• if w is a WFF, Nw is a WFF
• if w and x are WFFs, Kwx, Awx, Cwx, and Ewx are WFFs.

The meaning of a WFF is defined as follows:

• p, q, r, s, and t are logical variables that may take on the value 0 (false) or 1 (true).
• K, A, N, C, E mean and, or, not, implies, and equals as defined in the truth table below.

Definitions of K, A, N, C, and E     w  x  Kwx  Awx   Nw  Cwx  Ewx  1  1  1  1   0  1  1  1  0  0  1   0  0  0  0  1  0  1   1  1  0  0  0  0  0   1  1  1

A tautology is a WFF that has value 1 (true) regardless of the values of its variables. For example, ApNp is a tautology because it is true regardless of the value of p. On the other hand, ApNq is not, because it has the value 0 for p=0, q=1.

You must determine whether or not a WFF is a tautology.

Input

Input consists of several test cases. Each test case is a single line containing a WFF with no more than 100 symbols. A line containing 0 follows the last case.

Output

For each test case, output a line containing tautology or not as appropriate.

Sample Input

ApNpApNq0

Sample Output

tautologynot

判断一个表达式是否是重言式，所以用状压枚举所有情况，从后往前遍历字符串，用栈模拟

N !   A ||    K &&    C (!x)||y     E== 

#include <algorithm>#include <string.h>#include <iostream>#include <stdio.h>#include <string>#include <vector>#include <queue>#include <map>#include <set>#include <stack>using namespace std;typedef long long LL;const int N = 2010;int v[10];char str[100];stack<int>q;int main(){    while(scanf("%s",str),str[0]!='0')    {        int flag=1;        for(int i=0;i<(1<<5);i++)        {            memset(v,0,sizeof(v));            for(int j=0;j<5;j++)            {                if(i&(1<<j)) v[j]=1;            }            int len=strlen(str);            while(!q.empty()) q.pop();            for(int j=len-1;j>=0;j--)            {                if(str[j]>='p'&&str[j]<='t') q.push((v[str[j]-'p']));                else                {                    if(str[j]=='N')                    {                        int tmp=q.top();q.pop();                        q.push(!tmp);                        continue;                    }                    else if(str[j]=='A')                    {                        int tmp1=q.top();q.pop();                        int tmp2=q.top();q.pop();                        q.push(tmp1||tmp2);                        continue;                    }                    else if(str[j]=='K')                    {                        int tmp1=q.top();q.pop();                        int tmp2=q.top();q.pop();                        q.push(tmp1&&tmp2);                        continue;                    }                    else if(str[j]=='E')                    {                        int tmp1=q.top();q.pop();                        int tmp2=q.top();q.pop();                        q.push(tmp1==tmp2);                        continue;                    }                    else                    {                        int tmp1=q.top();q.pop();                        int tmp2=q.top();q.pop();                        q.push((!tmp1)||tmp2);                        continue;                    }                }            }            if(q.top()==0)            {                flag=0;                break;            }        }        if(flag) puts("tautology");        else puts("not");    }    return 0;}