SPOJ Classical problems 4 TRANSFORM THE EXPRESSION(ONP)

http://www.spoj.pl/problems/ONP/

Transform the algebraic expression with brackets into RPN form (Reverse Polish Notation). Two-argument operators: +, -, *, /, ^ (priority from the lowest to the highest), brackets ( ). Operands: only letters: a,b,...,z. Assume that there is only one RPN form (no expressions like a*b*c).

Input

t [the number of expressions <= 100]
expression [length <= 400]
[other expressions]

Text grouped in [ ] does not appear in the input file.

Output

The expressions in RPN form, one per line.

Example

Input:
3
(a+(b*c))
((a+b)*(z+x))
((a+t)*((b+(a+c))^(c+d)))

Output:
abc*+
ab+zx+*
at+bac++cd+^*

中缀表达式转后缀表达式，使用栈即可。

这里用stl库，不过自己手写效率要高一些。

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如为字母则放入结果

如为运算符，则将其与栈顶优先级比较，如低于栈顶则进栈，否则放入结果

左括号直接进栈，右括号退栈至最近的左括号

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#include <iostream>
#include <stack>
#include <cctype>
#include <vector>
#include <cstdio>
#include <string>

using namespace std;

char p[1000];

int main()
{
    freopen("in.txt", "r", stdin);



    p['+'] = '1';
    p['-'] = '2';
    p['*'] = '3';
    p['/'] = '4';
    p['^'] = '5';


    int n;
    cin >> n;
    while (n--)
    {
        stack<char> st;
        vector<char> cvec;
        char c;
        char priority = 6;
        string str;
        cin >> str;

        for (int i=0; i<str.size(); ++i)
        {
            c = str[i];
            if (isalpha(c))
            {
                cvec.push_back(c);
            }

            if (c == '(')
            {
                st.push(c);
                priority = '6';
            }
            if (c=='+' || c=='-' || c=='*' || c=='/' || c=='^')
            {
                if (p[c] >= priority)
                {
                    cvec.push_back(c);
                }
                else
                {
                    st.push(c);
                    priority = p[c];
                }
            }
            if (c == ')')
            {
                char retr = st.top();
                st.pop();

                while (retr != '(')
                {
                    cvec.push_back(retr);
                    retr = st.top();
                    st.pop();
                }
                priority = '6';
            }

        }
        for (vector<char>::iterator iter=cvec.begin(); iter!=cvec.end(); ++iter)
        {
            cout << *iter;
        }
        cout << endl;
    }
    return 0;
}