就不斷地掃整個序列，如果發現多余的括號就刪除。大概復雜度還是O(n2)左右。如何判斷不合法請詳見代碼。

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To a computer, there is no difference between the expression?(((x)+(y))(t))and?(x+y)t; but, to a human, the latter is easier to read. When writing automatically generated expressions that a human may have to read, it is useful to minimize the number of parentheses in an expression. We assume expressions consist of only two operations: addition (+) and multiplication (juxtaposition), and these operations act on single lower-case letter variables only. Specifically, here is the grammar for an expression?E:

E : P | P '+' E P : F | F P F : V | '(' E ')' V : 'a' | 'b' | .. | 'z'

The addition (+, as in?x+y) and multiplication (juxtaposition, as in?xy) operators are associative:?x+(y+z)=(x+y)+z=x+y+z?and?x(yz)=(xy)z=xyz. Commutativity and distributivity of these operations should not be assumed. Parentheses have the highest precedence, followed by multiplication and then addition.

Input

The input consists of a number of cases. Each case is given by one line that satisfies the grammar above. Each expression is at most 1000 characters long.

Output

For each case, print on one line the same expression with all unnecessary parentheses removed.

Sample input

x (x+(y+z)) (x+(yz)) (x+y(x+t)) x+y+xt

Sample Output

x x+y+z x+yz x+y(x+t) x+y+xt #include<cstdio> #include<iostream> #include<string> #include<stack> using namespace std; stack<int>st; string s; int n; bool check(int l,int r) {int cnt=0;for(int i=l;i<=r;++i)if(s[i]=='(')++cnt;else if(s[i]==')')--cnt;else if(s[i]=='+' && cnt==0)return 0;return 1; } int main() {while(cin>>s){while(1){bool flag=0;n=s.length();while(!st.empty())st.pop();for(int j=0;j<n;++j)if(s[j]=='(')st.push(j);else if(s[j]==')'){int x=st.top(); st.pop();if((x==0 || s[x-1]=='+' || s[x-1]=='(')&& (j==n-1 || s[j+1]=='+' || s[j+1]==')')){s.erase(x,1);s.erase(j-1,1);flag=1;break;}else if(check(x+1,j-1)){s.erase(x,1);s.erase(j-1,1);flag=1;break;}}if(!flag)break;}cout<<s<<endl;}return 0; }

轉載于:https://www.cnblogs.com/autsky-jadek/p/6291553.html

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