[ACM] POJ 3295 Tautology (构造）

Tautology

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 9302		Accepted: 3549

Description

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

ApNp
ApNq
0

Sample Output

tautology
not

Source

field=source&key=Waterloo+Local+Contest" style="text-decoration:none">Waterloo Local Contest, 2006.9.30

解题思路：

题意为依据输入的不同的字符串。来求这个逻辑表达式的值。

这里用到了栈，和表达式求值思想类似。从后往前扫面输入的字符串，假设是数（题中的p,q,r,s,t）就进栈。假设是操作符就从栈中取数。运算后再进栈。题中的数p,q,r,s,t仅仅有0,1取值，所以5重循环，对获得的表达式求值。

遇到0就退出。

代码：

#include <iostream>
#include <string.h>
#include <stack>
using namespace std;

string wff;
int p,q,r,s,t;
bool ok;

int compute(string str)//栈中的操作
{
    stack<int>st;
    int len=str.length();
    for(int i=len-1;i>=0;i--)
    {
        if(str[i]=='p')
            st.push(p);
        else if(str[i]=='q')
            st.push(q);
        else if(str[i]=='r')
            st.push(r);
        else if(str[i]=='s')
            st.push(s);
        else if(str[i]=='t')
            st.push(t);
        else if(str[i]=='K')
        {
            int x=st.top();
            st.pop();
            int y=st.top();
            st.pop();
            st.push(x&y);
        }
        else if(str[i]=='A')
        {
            int x=st.top();
            st.pop();
            int y=st.top();
            st.pop();
            st.push(x||y);
        }
        else if(str[i]=='N')
        {
            int x=st.top();
            st.pop();
            st.push(!x);
        }
        else if(str[i]=='C')
        {
            int x=st.top();
            st.pop();
            int y=st.top();
            st.pop();
            st.push(!x||y);
        }
        else if(str[i]=='E')
        {
            int x=st.top();
            st.pop();
            int y=st.top();
            st.pop();
            st.push(x==y);
        }
    }
    return st.top();
}

int main()
{
    while(cin>>wff&&wff!="0")
    {
        ok=1;
        for(p=0;p<2;p++)//枚举结果。遇到0就退出循环
            for(q=0;q<2;q++)
                for(r=0;r<2;r++)
                    for(s=0;s<2;s++)
                        for(t=0;t<2;t++)
        {
            if(compute(wff)==0)
            {
                ok=0;
                goto label;
            }
        }
        label:
        if(ok)
            cout<<"tautology"<<endl;
        else
            cout<<"not"<<endl;
    }
    return 0;
}