POJ3295

 转载：http://blog.csdn.net/AllenLSY/archive/2009/11/27/4885948.aspx

Tautology

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
0Sample Output

tautology
notSource

Waterloo Local Contest, 2006.9.30
总结：

递归题，表达式，永真式

 

AC代码1

 

01.#include <iostream> 02.#include <string> 03.#include <cctype> 04.using namespace std; 05.int p, q, r, s, t; 06.int K[2][2]={0, 0, 0, 1}, A[2][2]={0, 1, 1, 1}, N[2]={1, 0}, 07. C[2][2]={1, 1, 0, 1}, E[2][2]={1, 0, 0, 1}; 08.int getv(char ch) 09.{ 10. switch (ch) 11. { 12. case (int)'p': return p; 13. case (int)'q': return q; 14. case (int)'r': return r; 15. case (int)'s': return s; 16. case (int)'t': return t; 17. } 18.} 19.int dfs(string str, int& len) 20.{ 21. int a, b, ret, len1=0, len2=0; 22. if (isupper(str[1])) 23. a=dfs( str.substr(1), len1); 24. else 25. { 26. a=getv(str[1]); 27. len1=1; 28. } 29. if (str[0]!='N') 30. { 31. if (isupper(str[len1+1])) 32. b=dfs( str.substr(len1+1), len2 ); 33. else 34. { 35. b=getv(str[len1+1]); 36. len2=1; 37. } 38. } 39. len=1+len1+len2; 40. switch(str[0]) 41. { 42. case (int)'K': ret=K[a][b]; break; 43. case (int)'A': ret=A[a][b]; break; 44. case (int)'N': ret=N[a]; break; 45. case (int)'C': ret=C[a][b]; break; 46. case (int)'E': ret=E[a][b]; break; 47. } 48. return ret; 49.} 50.int main() 51.{ 52. string str; 53. while (cin >> str && str!="0") 54. { 55. p=0, q=0, r=0, s=0, t=0; 56. int len=0; 57. bool done=false; 58. for (p=0;p<=1 && !done;p++) 59. for (q=0;q<=1 && !done; q++) 60. for (r=0;r<=1 && !done; r++) 61. for (s=0;s<=1 && !done; s++) 62. for (t=0;t<=1 && !done; t++) 63. if (dfs(str, len)!=1) 64. done=true; 65. if (done) 66. cout << "not/n"; 67. else 68. cout << "tautology/n"; 69. } 70. return 0; 71.}

 

 

后来看了discuss的提示，给出了另一个很精简的写法

AC代码2

 

01.#include <iostream> 02.#include <string> 03.using namespace std; 04.int p, q, r, s, t; 05.int K[2][2]={0, 0, 0, 1}, A[2][2]={0, 1, 1, 1}, N[2]={1, 0}, 06. C[2][2]={1, 1, 0, 1}, E[2][2]={1, 0, 0, 1}; 07.string str; 08.int now; 09.int calc() 10.{ 11. now++; 12. switch(str[now]) 13. { 14. case (int)'K':return calc() & calc(); 15. case (int)'A':return calc() | calc(); 16. case (int)'N':return !calc(); 17. case (int)'C':return C[calc()][calc()]; 18. case (int)'E':return E[calc()][calc()]; 19. case (int)'p':return p; 20. case (int)'q':return q; 21. case (int)'r':return r; 22. case (int)'s':return s; 23. case (int)'t':return t; 24. } 25.} 26.int main() 27.{ 28. while (cin >> str && str!="0") 29. { 30. bool done=false; 31. for (p=0;p<=1 && !done;p++) 32. for (q=0;q<=1 && !done; q++) 33. for (r=0;r<=1 && !done; r++) 34. for (s=0;s<=1 && !done; s++) 35. for (t=0;t<=1 && !done; t++) 36. { 37. now=-1; 38. if (calc()!=1) 39. done=true; 40. } 41. if (done) 42. cout << "not/n"; 43. else 44. cout << "tautology/n"; 45. } 46. return 0; 47.}  

 

在网上，还发现一个用位运算的更精简的写法，学习了。

http://www.cnblogs.com/asuran/archive/2009/09/28/1575379.html

01.// POJ3295.cpp : Defines the entry point for the console application. 02.// 03.#include <iostream> 04.#include <string> 05.using namespace std; 06.static int pos = -1; 07.bool WFF(const string& formula, int i) 08.{ 09. ++pos; 10. switch(formula[pos]) 11. { 12. case 'p': 13. return i & 1; 14. case 'q': 15. return (i >> 1) & 1; 16. case 'r': 17. return (i >> 2) & 1; 18. case 's': 19. return (i >> 3) & 1; 20. case 't': 21. return (i >> 4) & 1; 22. case 'N': 23. return !WFF(formula, i); 24. case 'K': 25. return WFF(formula, i) & WFF(formula, i); 26. case 'A': 27. return WFF(formula, i) | WFF(formula, i); 28. case 'C': 29. return !WFF(formula, i) | WFF(formula, i); 30. case 'E': 31. return WFF(formula, i) == WFF(formula, i); 32. } 33. 34. return false; 35.}; 36.bool isTautology(string formula) 37.{ 38. for (int i = 0; i < 32; ++i) 39. { 40. pos = -1; 41. if (WFF(formula, i)==false) return false;; 42. } 43. return true; 44.}; 45.int main(int argc, char* argv[]) 46.{ 47. string ln; 48. while (cin >> ln && ln[0] != '0') 49. { 50. if (isTautology(ln)) cout << "tautology/n"; 51. else cout << "not/n"; 52. } 53. return 0; 54.}