POJ-1400（删除冗余括号）

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题目：http://poj.org/problem?id=1400

一开始在判断括号内是否全是乘或除时忘了括号内还有括号的情况，WA了两次之后找到了数据，对比答案之后才发现这个问题……删括号规则：

（1）括号前是除号，不能删

（2）括号前是乘号或减号，如果括号内全是乘除法或者被括号括起来的运算

（3）括号前是加法，如果括号后面是加法、括号后面是减法、括号内全是乘除法或者被括号括起来的运算

#include <string>#include <iostream>using namespace std;inline bool isOperator(char c){    return c == '+' || c == '-' || c == '*' || c == '/';}bool allMulOrDiv(const string& exp){    for(int i = 0, len = exp.size(); i < len; ++i){        if(exp[i] == '('){            int cnt = 1;            for(++i; true; ++i){                if(exp[i] == '(') ++cnt;                else if(exp[i] == ')'){                    --cnt;                    if(!cnt) break;                }            }            ++i;        }        if(exp[i] == '+' || exp[i] == '-') return false;    }    return true;}int findMatchedLeft(const string& exp, int r){    int countOfRight = 1;    for(--r; r >= 0; --r){        if(exp[r] == ')') ++countOfRight;        else if(exp[r] == '('){            --countOfRight;            if(!countOfRight) break;        }    }    return r;}string& simplify(string& exp){    int left, right = 0;    char head, tail, can;    while((right = exp.find(')', right)) != string::npos){        left = findMatchedLeft(exp, right);    //get info        if(left && isOperator(exp[left-1])) head = exp[left-1];        else head = '+';        if(right+1 < exp.size() && isOperator(exp[right+1])) tail = exp[right+1];        else tail = '+';    //analyze        if(right - left == 2) can = 1;        else if(head == '/') can = 0;        else if(head == '*' || head == '-') can = allMulOrDiv(exp.substr(left+1, right-left-1));        else can = tail == '+' || tail == '-' || allMulOrDiv(exp.substr(left+1, right-left-1));    //process        if(can){            exp.erase(left, 1).erase(right - 1, 1);            --right;        }        else ++right;    }    return exp;}int main(){    ios::sync_with_stdio(false);    int test;    cin >> test;    while(cin.get() != '\n') cin.get();    string exp;    while(test--){        getline(cin, exp);        cout << simplify(exp) << "\n";    }    return 0;}

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