• 《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 }
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  • 原文地址:https://www.cnblogs.com/zhuli19901106/p/3616671.html
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