POJ 3295 Tautology(构造法)

Posted 2020-09-29 ventricle

题目网址：http://poj.org/problem?id=3295

题目：

Tautology

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 13231		Accepted: 5050

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

思路：
枚举 p, q, r, s, t的所有情况，共2^5种情况，对于每种情况都对给定的字符串进行操作，只要有一种情况是假的，结果就是假的，反之则是永真。
字符串长度不会超过100，所以最坏的情况是2^5*100 肯定不会超时。

对字符串的处理：从字符串尾部进行判断，如果是p,q,r,s,t就压入栈，如果是逻辑符号，就取栈顶元素运算。

代码：

 1 #include <cstdio>
 2 #include <stack>
 3 #include <cstring>
 4 using namespace std;
 5 char str[105];
 6 int p,q,r,s,t;
 7 int solve(){
 8     stack<int>num;
 9     int len=strlen(str);
10     for (int i=len-1; i>=0; i--) {
11         if(str[i]>‘a‘ && str[i]<‘z‘){
12             if(str[i]==‘p‘) num.push(p);
13             if(str[i]==‘q‘) num.push(q);
14             if(str[i]==‘r‘) num.push(r);
15             if(str[i]==‘s‘) num.push(s);
16             if(str[i]==‘t‘) num.push(t);
17         }else{
18             int a=num.top();num.pop();
19             if(str[i]==‘N‘) num.push(!a);
20             else{
21                 int b=num.top();num.pop();
22                 if(str[i]==‘K‘) num.push(a&b);
23                 if(str[i]==‘A‘) num.push(a|b);
24                 if(str[i]==‘C‘) num.push((!a)|b);
25                 if(str[i]==‘E‘) num.push(a==b?1:0);
26             }
27         }
28     }
29     return num.top();
30 }
31 int main() {
32     while (gets(str)!=NULL && str[0]!=‘0‘) {
33         int flag=0;
34         for (p=0; p<2 ; p++) {
35             for (q=0; q<2; q++) {
36                 for (r=0; r<2; r++) {
37                     for (s=0; s<2; s++) {
38                         for (t=0; t<2; t++) {
39                             if(solve()==0){
40                                 flag=1;
41                                 break;
42                             }
43                         }
44                     }
45                 }
46             }
47         }
48         if(flag)    printf("not\n");
49         else printf("tautology\n");
50     }
51     return 0;
52 }