表達式得到期望結果的組成種數

/************************************************************************************
 *description:表達式得到期望結果的組成種數
 *            給定一個字符串express,只由false(假)、true(真)、&(與)、|(或)、^(異或)
 *            這5種字符組成。再給定一個bool值desired。
 *            返回express能有多少種組合方式達到desired
 *            例：express=“1^0|0|1",desired=false
 *                1^((0|0)|1)和1^(0|(0|1))可以得到false，返回2

 ***********************************************************************************/

#include<iostream>
#include<vector>
#include<string>
using namespace std;
//有效的表達式：1.長度爲奇數
//            2.下標偶數位的字符是：0或1
//            3.下標奇數爲的字符是：&、|、^

bool isValid(string express)
{
    if (express.size() % 2 == 0)
        return false;
    for (int i = 0; i < express.size(); i+=2)
        if (express[i] == '&' || express[i] == '|' || express[i] == '^')
            return false;
    for (int i = 1; i < express.size(); i+=2)
        if (express[i] == '0' || express[i] == '1')
            return false;
    return true;

}

//方法1：暴力遞歸.時間複雜度O(N!),空間複雜度O(N)：函數棧大小
int process_num1(string exp, bool desired, int left, int right)
{
    if (left == right)
        if (exp[left] == '1')
            return desired ? 1 : 0;
        else
            return desired ? 0 : 1;
    int res = 0;
    for (int i = left+1; i < right; i+=2)
    {
        if (desired == true)
        {
            switch (exp[i])
            {
            case '&':
                res += process_num1(exp, true, left, i-1) * process_num1(exp, true, i+1, right);
                break;
            case '|':
                res += process_num1(exp, true, left, i-1) * process_num1(exp, true, i+1, right)
                        +process_num1(exp, true, left, i-1) * process_num1(exp, false, i+1, right)
                        +process_num1(exp, false, left, i-1) * process_num1(exp, true, i+1, right);
                break;
            case '^':
                res += process_num1(exp, true, left, i-1) * process_num1(exp, false, i+1, right)
                        +process_num1(exp, false, left, i-1) * process_num1(exp, true, i+1, right);
            default:
                break;
            }
        }
        else
        {
            switch (exp[i])
            {
            case '&':
                res += process_num1(exp, true, left, i-1) * process_num1(exp, false, i+1, right)
                        +process_num1(exp, false, left, i-1) * process_num1(exp, true, i+1, right)
                        +process_num1(exp, false, left, i-1) * process_num1(exp, false, i+1, right);
                break;
            case '|':
                res += process_num1(exp, false, left, i-1) * process_num1(exp, false, i+1, right);
                break;
            case '^':
                res += process_num1(exp, true, left, i-1) * process_num1(exp, true, i+1, right)
                        +process_num1(exp, false, left, i-1) * process_num1(exp, false, i+1, right);
            default:
                break;
            }
        }
    }
    return res;
}
int numOfCombine_1(string express, bool desired)
{
    if (express.size() == 0)
        return 0;
    if (isValid(express) == false)
        return 0;
    return process_num1(express, desired, 0, express.size()-1);
}
//方法2：動態規劃.時間複雜度O(N^3),空間複雜度O(N^2)
//建立數組 t[N][N]和f[N][N],其中N爲express的大小。
//t[j][i]表示express[j...i]組成true的種類；f[j][i]表示express[j...i]組成false的種類。
//t[0][N-1]或f[0][N-1]即爲結果。
int numOfCombine_2(string express, bool desired)
{
    if (express.size() == 0)
        return 0;
    if (isValid(express) == false)
        return 0;
    int N = express.size();
    vector<vector<int>> t(N, N);
    vector<vector<int>> f(N, N);

    t[0][0] = express[0] == '1' ? 1 : 0;
    f[0][0] = express[0] == '0' ? 1 : 0;

    for (int i = 2; i < N; i += 2 )
    {
        t[i][i] = express[i] == '1' ? 1 : 0;
        f[i][i] = express[i] == '0' ? 1 : 0;
        for (int j = i - 2; j >= 0; j -= 2)
        {
            for (int k = j; k < i; k += 2)
            {
                if (express[k+1] == '&')
                {
                    t[j][i] += t[j][k] * t[k+2][i];
                    f[i][j] += (f[j][k] + t[j][k])*f[k+2][i]+f[j][k]*t[k+2][i];
                }
                else if (express[k+1] == '|')
                {
                    t[j][i] += (f[j][k] + t[j][k])*t[k+2][i]+t[j][k]*f[k+2][i];
                    f[j][i] += f[j][k] * f[k+2][i];
                }
                else
                {
                    t[j][i] = f[j][k] * t[k+2][i] + t[j][k] * f[k+2][i];
                    f[j][i] = f[j][k] * f[k+2][i] + t[j][k] * t[k+2][i];
                }
            }
        }
    }
    return desired == true ? t[0][N-1] : f[0][N-1];
}

int main()
{
    string exp = "1^0|0|1";
    bool desired = false;
    cout << numOfCombine_2(exp, desired);


    return 0;
}