《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() == 1) {
        if (exp[0] - '0' == target) {
            path.push_back(one_path);
            ++result;
        }
        return;
    }
    
    string new_exp;
    int i, j;
    for (i = 1; i < (int)exp.length(); i += 2) {
        new_exp = "";
        
        for (j = 0; j < i - 1; ++j) {
            new_exp.push_back(exp[j]);
        }
        
        if (exp[i] == '&') {
            new_exp.push_back(((exp[i - 1] - '0') & (exp[i + 1] - '0')) + '0');
        } else if (exp[i] == '|') {
            new_exp.push_back(((exp[i - 1] - '0') | (exp[i + 1] - '0')) + '0');
        } else if (exp[i] == '^') {
            new_exp.push_back(((exp[i - 1] - '0') ^ (exp[i + 1] - '0')) + '0');
        }
        
        for (j = i + 2; 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 = 0;
        one_path.push_back(exp);
        booleanExpressionParentheses(exp, target, result, one_path, path);
        one_path.pop_back();
        for (i = 0; i < (int)path.size(); ++i) {
            cout << "Path " << i + 1 << endl;
            for (j = 0; j < (int)path[i].size(); ++j) {
                cout << path[i][j] << endl;
            }
            cout << endl;
        }

        for (i = 0; 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 0;
}