Complicated Expressions(表达式转换)

http://poj.org/problem?id=1400

题意：给出一个表达式可能含有多余的括号，去掉多余的括号，输出它的最简形式。

思路：先将表达式转化成后缀式，因为后缀式不含括号，然后再转化成中缀式，并根据运算符的优先级添加括号。

 #include <stdio.h>
 #include <string.h>
 #include <string>
 #include <stack>
 #include <iostream>
 using namespace std;
 char ss[];
 int f(char ch)
 {
     if(ch=='(')
         return ;
     else if(ch=='+'||ch=='-')
         return ;
     else if (ch=='*'||ch=='/')
         return ;
 }
 void change1(string s)
 {
     int l = ;
     stack<char>p;
     while(!p.empty()) p.pop();
     for (int i = ; i < s.size(); i++)
     {
         if(s[i]>='a'&&s[i]<='z')
             ss[l++]=s[i];
         else
         {
             if (s[i]=='(')
                 p.push(s[i]);
             else if (s[i]==')')
             {
                 while(!p.empty())
                 {
                     char ch = p.top();
                     p.pop();
                     if(ch=='(')
                         break;
                     ss[l++] = ch;
                 }
             }
             else
             {
                 while(!p.empty()&&f(p.top())>=f(s[i]))
                 {
                     char ch = p.top();
                     p.pop();
                     ss[l++] = ch;
                 }
                 p.push(s[i]);
             }

         }
     }
 }
 while(!p.empty())
 {
     ss[l++] = p.top();
     p.pop();
 }
 ss[l++]='\0';
 }
 void change2(char a[])
 {
     string s1, s2, s3, fl, fs;
     string s[];
     char f[];
     int top = ;
     s1 = a;
     for(int i = ; i < (int)s1.length(); i++)
     {
         if(s1[i]>='a' && s1[i]<='z')
         {
             fl = s1[i];
             top++;
             s[top] = s1[i];
         }
         else if(s1[i]=='+')
         {
             s2 = s[top];
             top--;
             s3 = s[top];
             top--;
             top++;
             s[top] = s3+'+'+s2;
             f[top] = '+';
         }
         else if(s1[i]=='-')
         {
             s2 = s[top];
             top--;
             s3 = s[top];
             top--;
             if(f[top+]=='+' || f[top+]=='-')
             {
                 if(s2.length()>) s2 = '('+s2+')';
             }
             top++;
             s[top] = s3+'-'+s2;
             f[top] = '-';
         }
         else if(s1[i]=='*')
         {
             s2 = s[top];
             top--;
             s3 = s[top];
             top--;
             if(f[top+]=='+' || f[top+]=='-')
             {
                 if(s3.length()>) s3 = '('+s3+')';
             }
             if(f[top+]=='+' || f[top+]=='-')
             {
                 if(s2.length()>) s2 = '('+s2+')';
             }
             top++;
             s[top] = s3+'*'+s2;
             f[top] = '*';
         }
         else if(s1[i]=='/')
         {
             s2 = s[top];
             top--;
             s3 = s[top];
             top--;
             if(f[top+]=='+' || f[top+]=='-')
             {
                 if(s3.length()>) s3 = '('+s3+')';
             }
             if(f[top+]=='+' || f[top+]=='-' || f[top+]=='*' || f[top+]=='/')
             {
                 if(s2.length()>) s2='('+s2+')';
             }
             top++;
             s[top] = s3+'/'+s2;
             f[top] = '/';
         }
     }
     cout<<s[]<<endl;
 }
 int main()
 {

     //freopen("expr.in", "r", stdin);
     //freopen("ss.out", "w", stdout);
     int t;
     string s;
     cin>>t;
     while(t--)
     {
         cin>>s;
         change1(s);
         change2(ss);
     }
     return ;
 }