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。
1 #include <stdio.h> 2 #include <string.h> 3 #include <string> 4 #include <stack> 5 #include <iostream> 6 using namespace std; 7 char ss[1010]; 8 int f(char ch) 9 { 10 if(ch=='(') 11 return 0; 12 else if(ch=='+'||ch=='-') 13 return 1; 14 else if (ch=='*'||ch=='/') 15 return 2; 16 } 17 void change1(string s) 18 { 19 int l = 0; 20 stack<char>p; 21 while(!p.empty()) p.pop(); 22 for (int i = 0; i < s.size(); i++) 23 { 24 if(s[i]>='a'&&s[i]<='z') 25 ss[l++]=s[i]; 26 else 27 { 28 if (s[i]=='(') 29 p.push(s[i]); 30 else if (s[i]==')') 31 { 32 while(!p.empty()) 33 { 34 char ch = p.top(); 35 p.pop(); 36 if(ch=='(') 37 break; 38 ss[l++] = ch; 39 } 40 } 41 else 42 { 43 while(!p.empty()&&f(p.top())>=f(s[i])) 44 { 45 char ch = p.top(); 46 p.pop(); 47 ss[l++] = ch; 48 } 49 p.push(s[i]); 50 } 51 52 } 53 } 54 } 55 while(!p.empty()) 56 { 57 ss[l++] = p.top(); 58 p.pop(); 59 } 60 ss[l++]=' '; 61 } 62 void change2(char a[]) 63 { 64 string s1, s2, s3, fl, fs; 65 string s[300]; 66 char f[300]; 67 int top = 0; 68 s1 = a; 69 for(int i = 0; i < (int)s1.length(); i++) 70 { 71 if(s1[i]>='a' && s1[i]<='z') 72 { 73 fl = s1[i]; 74 top++; 75 s[top] = s1[i]; 76 } 77 else if(s1[i]=='+') 78 { 79 s2 = s[top]; 80 top--; 81 s3 = s[top]; 82 top--; 83 top++; 84 s[top] = s3+'+'+s2; 85 f[top] = '+'; 86 } 87 else if(s1[i]=='-') 88 { 89 s2 = s[top]; 90 top--; 91 s3 = s[top]; 92 top--; 93 if(f[top+2]=='+' || f[top+2]=='-') 94 { 95 if(s2.length()>1) s2 = '('+s2+')'; 96 } 97 top++; 98 s[top] = s3+'-'+s2; 99 f[top] = '-'; 100 } 101 else if(s1[i]=='*') 102 { 103 s2 = s[top]; 104 top--; 105 s3 = s[top]; 106 top--; 107 if(f[top+1]=='+' || f[top+1]=='-') 108 { 109 if(s3.length()>1) s3 = '('+s3+')'; 110 } 111 if(f[top+2]=='+' || f[top+2]=='-') 112 { 113 if(s2.length()>1) s2 = '('+s2+')'; 114 } 115 top++; 116 s[top] = s3+'*'+s2; 117 f[top] = '*'; 118 } 119 else if(s1[i]=='/') 120 { 121 s2 = s[top]; 122 top--; 123 s3 = s[top]; 124 top--; 125 if(f[top+1]=='+' || f[top+1]=='-') 126 { 127 if(s3.length()>1) s3 = '('+s3+')'; 128 } 129 if(f[top+2]=='+' || f[top+2]=='-' || f[top+2]=='*' || f[top+2]=='/') 130 { 131 if(s2.length()>1) s2='('+s2+')'; 132 } 133 top++; 134 s[top] = s3+'/'+s2; 135 f[top] = '/'; 136 } 137 } 138 cout<<s[1]<<endl; 139 } 140 int main() 141 { 142 143 //freopen("expr.in", "r", stdin); 144 //freopen("ss.out", "w", stdout); 145 int t; 146 string s; 147 cin>>t; 148 while(t--) 149 { 150 cin>>s; 151 change1(s); 152 change2(ss); 153 } 154 return 0; 155 }