《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：

 // 9.11 For a boolean expression and a target value, calculate how many ways to use parentheses on the expression to achieve that value.
 #include <iostream>
 #include <string>
 #include <vector>
 using namespace std;

 void booleanExpressionParentheses(string exp, int target, int &result, vector<string> &one_path, vector<vector<string> > &path)
 {
     if ((int)exp.length() == ) {
         if (exp[] - '' == target) {
             path.push_back(one_path);
             ++result;
         }
         return;
     }

     string new_exp;
     int i, j;
     for (i = ; i < (int)exp.length(); i += ) {
         new_exp = "";

         for (j = ; j < i - ; ++j) {
             new_exp.push_back(exp[j]);
         }

         if (exp[i] == '&') {
             new_exp.push_back(((exp[i - ] - '') & (exp[i + ] - '')) + '');
         } else if (exp[i] == '|') {
             new_exp.push_back(((exp[i - ] - '') | (exp[i + ] - '')) + '');
         } else if (exp[i] == '^') {
             new_exp.push_back(((exp[i - ] - '') ^ (exp[i + ] - '')) + '');
         }

         for (j = i + ; j < (int)exp.length(); ++j) {
             new_exp.push_back(exp[j]);
         }
         one_path.push_back(new_exp);
         booleanExpressionParentheses(new_exp, target, result, one_path, path);
         one_path.pop_back();
     }
 }

 int main()
 {
     int result;
     int target;
     int i, j;
     string exp;
     vector<vector<string> > path;
     vector<string> one_path;

     while (cin >> exp >> target) {
         result = ;
         one_path.push_back(exp);
         booleanExpressionParentheses(exp, target, result, one_path, path);
         one_path.pop_back();
         for (i = ; i < (int)path.size(); ++i) {
             cout << "Path " << i +  << endl;
             for (j = ; j < (int)path[i].size(); ++j) {
                 cout << path[i][j] << endl;
             }
             cout << endl;
         }

         for (i = ; i < (int)path.size(); ++i) {
             path[i].clear();
         }
         path.clear();
         one_path.clear();

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

     return ;
 }