Tautology-状压

Tautology

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 11716		Accepted: 4437

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

题目链接：http://poj.org/problem?id=3295

题目大意：

此题是让你判断式子最后的答案是否永真，离散好的大神应该看图就能看懂题，根据上表，判断输入的字符串是否为永真式，poj水题，本想不懂脑子用多重for循环直接水过，但是恰巧金桔路过，深深地鄙视了我的for循环，我。。。。。。最终决定，上状压，当然，金桔就是金桔，代码力求简洁。

题解：

共五个变元，p,q,r,s,t，每一个变元有0和1两种状态，共有1<<5种状态，对应每一个变元的值，直接上状压！

本渣的代码：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <stack>
using namespace std;
bool f(int h1,int h2,char c)
{
switch(c)
{
case 'K':
return h1&&h2;
case 'A':
return h1||h2;
case 'C':
return (!h1&&h2)||h1==h2;
case 'E':
return h1==h2;
default :
break;
}
return 0;
}
bool judge(char s[])
{
int len=strlen(s);
bool v[10];
stack<bool> Q;
int i,j;
for(i=0;i<1<<5;i++)
{
int t=i;
for(j=0;j<5;j++)
{
v[j]=t%2;
t>>=1;
}
while(!Q.empty())
Q.pop();
for(j=len-1;j>=0;j--)
{
if(s[j]>='p'&&s[j]<='t')
Q.push(v[s[j]-'p']);
else if(s[j]=='N')
{
int h=Q.top();
Q.pop();
h=!h;
Q.push(h);
}
else
{
int h1=Q.top();
Q.pop();
int h2=Q.top();
Q.pop();
int h=f(h1,h2,s[j]);
Q.push(h);
}
}
if(!Q.top())
return 0;
}
if(Q.top())
return 1;
return 0;
}
int main()
{
char s[1000];
while(~scanf("%s",s)&&*s!='0')
{
printf(judge(s)?"tautology\n":"not\n");
}
return 0;
}

好久没有写点东西了，颓废了好长时间了，也落后了别人好多了，现在只想认真努力，借用前两天看的电影中的一句话，我们学习不是为了分数和比赛的成绩，而是为了完善我们自己，加油吧！