《Cracking the Coding Interview》——第9章：递归和动态规划——题目11

2014-03-21 20:20

题目：给定一个只包含‘0’、‘1’、‘|’、‘&’、‘^’的布尔表达式，和一个期望的结果(0或者1)。如果允许你用自由地给这个表达式加括号来控制运算的顺序，问问有多少种加括号的方法来达到期望的结果值。

解法：DFS暴力解决，至于优化方法，应该是可以进行一部分剪枝的，但我没想出能明显降低复杂度的方法。

代码：

 1 // 9.11 For a boolean expression and a target value, calculate how many ways to use parentheses on the expression to achieve that value.

 2 #include <iostream>

 3 #include <string>

 4 #include <vector>

 5 using namespace std;

 6 

 7 void booleanExpressionParentheses(string exp, int target, int &result, vector<string> &one_path, vector<vector<string> > &path)

 8 {

 9     if ((int)exp.length() == 1) {

10         if (exp[0] - '0' == target) {

11             path.push_back(one_path);

12             ++result;

13         }

14         return;

15     }

16     

17     string new_exp;

18     int i, j;

19     for (i = 1; i < (int)exp.length(); i += 2) {

20         new_exp = "";

21         

22         for (j = 0; j < i - 1; ++j) {

23             new_exp.push_back(exp[j]);

24         }

25         

26         if (exp[i] == '&') {

27             new_exp.push_back(((exp[i - 1] - '0') & (exp[i + 1] - '0')) + '0');

28         } else if (exp[i] == '|') {

29             new_exp.push_back(((exp[i - 1] - '0') | (exp[i + 1] - '0')) + '0');

30         } else if (exp[i] == '^') {

31             new_exp.push_back(((exp[i - 1] - '0') ^ (exp[i + 1] - '0')) + '0');

32         }

33         

34         for (j = i + 2; j < (int)exp.length(); ++j) {

35             new_exp.push_back(exp[j]);

36         }

37         one_path.push_back(new_exp);

38         booleanExpressionParentheses(new_exp, target, result, one_path, path);

39         one_path.pop_back();

40     }

41 }

42 

43 int main()

44 {

45     int result;

46     int target;

47     int i, j;

48     string exp;

49     vector<vector<string> > path;

50     vector<string> one_path;

51     

52     while (cin >> exp >> target) {

53         result = 0;

54         one_path.push_back(exp);

55         booleanExpressionParentheses(exp, target, result, one_path, path);

56         one_path.pop_back();

57         for (i = 0; i < (int)path.size(); ++i) {

58             cout << "Path " << i + 1 << endl;

59             for (j = 0; j < (int)path[i].size(); ++j) {

60                 cout << path[i][j] << endl;

61             }

62             cout << endl;

63         }

64 

65         for (i = 0; i < (int)path.size(); ++i) {

66             path[i].clear();

67         }

68         path.clear();

69         one_path.clear();

70 

71         cout << "Total number of ways to achieve the target value is " << result << "." << endl;

72     }

73     

74     return 0;

75 }