HDU 5479 Scaena Felix

Scaena Felix

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 258    Accepted Submission(s): 112

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

102

BC，今天直接哭了，本来A了3道，坐等涨分，结果醒来一看，全卡掉了。。。。

不过，今天搞到女神的照片了，哈哈哈

#include <iostream>#include <string>#include <stdlib.h>#include <ctype.h>#include <cstdio>#include <cstdlib>#include <set>#include <map>#include <queue>#include <stack>#include <cstring>#include <math.h>#include <algorithm>#define LL long long#define INF 0x3f3f3f3f#define RR freopen("in.txt","r",stdin)#define WW freopen("out.txt","w",stdout)#define PI acos(-1.0)using namespace std;int L[1010],R[1010];char s[1010];int main(){    int T;    cin>>T;    while(T--)    {        cin>>s+1;        int len = strlen(s+1);        for(int i=1;i<=len;i++)        {            L[i] = L[i-1] + (s[i] == '(');        }        for(int i=len;i>=1;i--)        {            R[i] = R[i+1] + (s[i] == ')');        }        int ans = INF;        for(int i=0;i<=len;i++)        {            ans = min(ans,L[i]+R[i+1]);        }        cout<<ans<<endl;    }    return 0;}