Coderforce 380 C. Sereja and Brackets(线段树)

C. Sereja and Brackets

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

Sereja has a bracket sequence s1, s2, ..., sn,
or, in other words, a string s of length n,
consisting of characters "(" and ")".

Sereja needs to answer m queries, each of them is described by two integers li, ri (1 ≤ li ≤ ri ≤ n).
The answer to the i-th query is the length of the maximum correct bracket subsequence of sequence sli, sli + 1, ..., sri.
Help Sereja answer all queries.

You can find the definitions for a subsequence and a correct bracket sequence in the notes.

Input

The first line contains a sequence of characters s1, s2, ..., sn (1 ≤ n ≤ 106) without
any spaces. Each character is either a "(" or a ")".
The second line contains integer m (1 ≤ m ≤ 105) —
the number of queries. Each of the next m lines contains a pair of integers. The i-th
line contains integers li, ri (1 ≤ li ≤ ri ≤ n) —
the description of the i-th query.

Output

Print the answer to each question on a single line. Print the answers in the order they go in the input.

Sample test(s)

input

())(())(())(
7
1 1
2 3
1 2
1 12
8 12
5 11
2 10

output

0
0
2
10
4
6
6

Note

A subsequence of length |x| of string s = s1s2... s|s| (where |s| is
the length of string s) is string x = sk1sk2... sk|x|(1 ≤ k1 < k2 < ... < k|x| ≤ |s|).

A correct bracket sequence is a bracket sequence that can be transformed into a correct aryphmetic expression by inserting characters "1"
and "+" between the characters of the string. For example, bracket sequences "()()",
"(())" are correct (the resulting expressions "(1)+(1)",
"((1+1)+1)"), and ")(" and "("
are not.

For the third query required sequence will be «()».

For the fourth query required sequence will be «()(())(())».

题目大意：一串括号，区间里边有多少个括号是匹配的

思路：http://blog.csdn.net/keshuai19940722/article/details/19014821

首先遍历一遍，将每个位置的从1到当前位置的有效右括号数和没有用到的左括号数记录下来；然后每次查询a，b区间，即为t = r[b] - r[a-1](有效右括号数），但是要注意，这些有效右括号的匹配左括号可能不在区间上，所以要减去l[a-1]（未用到的左括号数），但是又有可能[1,a-1]中未用到的左括号和[b+1,len]中的右括号相匹配，所以要加上min(l[a~b]),这不用到线段树查询最小值。

#	Author	Problem	Lang	Verdict	Time	Memory	Sent	Judged
14507935	Practice: kxh1995	380C - 18	MS C++	Accepted	139 ms	24468 KB	2015-11-27 11:35:48	2015-11-27 11:35:48

#include<stdio.h>
#include<stdlib.h>
#include<iostream>
#include<algorithm>
#include<string.h>
#define N 1000010
using namespace std;
char str
;
int LL
,RR
;
int node[N<<2];
void pushup(int tr)
{
node=min(node[tr<<1],node[tr<<1|1]);
}
void build(int l,int r,int tr)
{
if(l==r)
{
node=LL[l];
return;
}
int mid=(l+r)>>1;
build(l,mid,tr<<1);
build(mid+1,r,tr<<1|1);
pushup(tr);
}
int query(int L,int R,int l,int r,int tr)
{
if(L<=l&&r<=R)
{
return node;
}
int mid=(l+r)>>1;
if(R<=mid)
return query(L,R,l,mid,tr<<1);
else
if(L>mid)
return query(L,R,mid+1,r,tr<<1|1);
else
{
int a=query(L,mid,l,mid,tr<<1);
int b=query(mid+1,R,mid+1,r,tr<<1|1);
return min(a,b);
}
}
int main()
{
while(scanf("%s",str+1)!=EOF)
{
int q;
scanf("%d",&q);
int i;
int n=strlen(str+1);
int temp=0;
memset(LL,0,sizeof(LL));
memset(RR,0,sizeof(RR));
for(i=1;i<=n;i++)
{
RR[i]=RR[i-1];
if(str[i]=='(')
temp++;
else
{
RR[i]++;
temp--;
}
LL[i]=temp;
}
build(1,n,1);
while(q--)
{
int a,b;
scanf("%d%d",&a,&b);
/*if(a==b)
{
printf("0\n");
continue;
}*/
printf("%d\n",(RR[b]-RR[a-1]-max(LL[a-1]-query(a,b,1,n,1),0))*2);
}
}
}