Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 492 Accepted Submission(s): 214
Problem Description
Given a parentheses sequence consist of ‘(‘ and ‘)‘, a modify can filp a parentheses, changing ‘(‘ to ‘)‘ or ‘)‘ to ‘(‘.
If we want every not empty <b>substring</b> of this parentheses sequence not to be "paren-matching", how many times at least to modify this parentheses sequence?
For example, "()","(())","()()" are "paren-matching" strings, but "((", ")(", "((()" are not.
Input
The first line of the input is a integer T, meaning that there are T test cases.
Every test cases contains a parentheses sequence S only consists of ‘(‘ and ‘)‘.
1≤|S|≤1,000.
Output
For every test case output the least number of modification.
Sample Input
3 () (((( (())
Sample Output
1 0 2
题目最终就是要要求统计的匹配括号对，实际就是栈的应用。
源码：
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<string>
#include<stack>
#include<vector>
using namespace std;
#define MAX 1009
template <class Type>
class Stack {
public:
Stack(int MaxStackSize = MAX);
bool IsFull();
bool IsEmpty();
void Push(const Type x);
Type Pop();
private:
int top;
Type *stack;
int MaxSize;
};
template<class Type>
Stack<Type>::Stack(int MaxStackSize) :MaxSize(MaxStackSize)
{
stack = new Type[MaxSize];
top = -1;
}
template<class Type>
inline bool Stack<Type>::IsFull()
{
if (top == MaxSize - 1)
return true;
else
return false;
}
template<class Type>
inline bool Stack<Type>::IsEmpty()
{
if (top == -1)
return true;
else
return false;
}
template<class Type>
void Stack<Type>::Push(const Type x)
{
if (IsFull())
;
else
stack[++top] = x;
}
template<class Type>
Type Stack<Type>::Pop()
{
if (IsEmpty())
{
return 0;
}
else
{
Type x = stack[top--];
return x;
}
return 0;
}
template<class Type>
int GetNum(const Type *S,int n)
{
int count = 0;
Type x = ‘ ‘;
Stack<Type> ST;
ST.Push(*S++);
for (int i = 1; i < n; i++)
{
x = ‘ ‘;
if (!ST.IsEmpty())
{
x = ST.Pop();
}
if (x == ‘(‘&&*S == ‘)‘)
{
count++;
*S++;
}
else
{
if (x!=‘ ‘)
ST.Push(x);
ST.Push(*S++);
}
}
return count;
}
int main()
{
int T = 0;
string S;
while (cin >> T) {
for (int i = T; i >= 1; i--)
{
cin >> S;
cout << GetNum(S.c_str(),S.size()) << endl;
}
}
return 0;
}
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原文地址：http://592669550.blog.51cto.com/5487419/1699918
