hdu5479Scaena Felix(好题)

Scaena Felix

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 105    Accepted Submission(s): 52

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
  

 

Source

BestCoder Round #57 (div.2)

/*
题目要求每一个子串都不是括号匹配串，
实际上只要使得括号匹配的单元“()”串不是该串的子串即可。
所以最终串肯定是连续数个')'
以及连续数个'('，统计出其中代价最小的即可。
*/
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <cmath>
#include <ctime>
#include <vector>
#include <cstdio>
#include <cctype>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
using namespace std;
#define INF 0x3f3f3f3f
#define inf -0x3f3f3f3f
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define mem0(a) memset(a,0,sizeof(a))
#define mem1(a) memset(a,-1,sizeof(a))
#define mem(a, b) memset(a, b, sizeof(a))
typedef long long ll;
char s[1010];
int l[1010],r[1010];

int main(){
    int t;
    scanf("%d",&t);
    while(t--){
        scanf("%s",s);
        mem0(l);
        mem0(r);
        int len=strlen(s);
        for(int i=len-1;i>=0;i--){
            if(i==len-1)
                r[i]=(s[i]==')');
            else
                r[i]=r[i+1]+(s[i]==')');
        }
        for(int i=0;i<len;i++){
            if(i==0)
                l[i]=(s[i]=='(');       //把左边的全部换成')'
            else
                l[i]=l[i-1]+(s[i]=='(');
        }
        int ans=INF;
        for(int i=0;i<len;i++)
            ans=min(ans,l[i]+r[i]-1);   //有一个字符同时算了两次
        printf("%d\n",ans);
    }
    return 0;
}