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。

================================================

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

=================================================

 1 #include <iostream>

 2 #include <stack>

 3 #include <cctype>

 4 #include <vector>

 5 #include <cstdio>

 6 #include <string>

 7 

 8 using namespace std;

 9 

10 char p[1000];

11 

12 int main()

13 {

14     freopen("in.txt", "r", stdin);

15 

16 

17 

18     p['+'] = '1';

19     p['-'] = '2';

20     p['*'] = '3';

21     p['/'] = '4';

22     p['^'] = '5';

23 

24 

25     int n;

26     cin >> n;

27     while (n--)

28     {

29         stack<char> st;

30         vector<char> cvec;

31         char c;

32         char priority = 6;

33         string str;

34         cin >> str;

35 

36         for (int i=0; i<str.size(); ++i)

37         {

38             c = str[i];

39             if (isalpha(c))

40             {

41                 cvec.push_back(c);

42             }

43 

44             if (c == '(')

45             {

46                 st.push(c);

47                 priority = '6';

48             }

49             if (c=='+' || c=='-' || c=='*' || c=='/' || c=='^')

50             {

51                 if (p[c] >= priority)

52                 {

53                     cvec.push_back(c);

54                 }

55                 else

56                 {

57                     st.push(c);

58                     priority = p[c];

59                 }

60             }

61             if (c == ')')

62             {

63                 char retr = st.top();

64                 st.pop();

65 

66                 while (retr != '(')

67                 {

68                     cvec.push_back(retr);

69                     retr = st.top();

70                     st.pop();

71                 }

72                 priority = '6';

73             }

74 

75         }

76         for (vector<char>::iterator iter=cvec.begin(); iter!=cvec.end(); ++iter)

77         {

78             cout << *iter;

79         }

80         cout << endl;

81     }

82     return 0;

83 }