Codeforces generation C. sereja and brackets [Line Segment tree]

The following figure shows how many matching brackets exist between [a, B] in M queries (including a, B: process Two arrays R [I] (representing the number of matched right brackets between 1 and I ), L [I] (represents the number of unmatched left parentheses between 1 and I ). We need to calculate the number of matching parentheses between [a, B], first use R [B]-R [A-1], but masking is more likely that the right brackets between A and B are matched between [1, A-1], then minus L [A-1], but minus this part, it is also possible that it is matched in the range [B + 1, Len]. To add the value that should not be subtracted from this part, you only need to add min (L [a] ~ L [B. If the interval is the minimum value, use the line segment tree.
Code:

#include <iostream>#include <cstdio>#include <cstring>#define N 1000010using namespace std;#define L(x) (x)<<1#define R(x) (x)<<1|1int m,len,r[N],l[N];char bra[N];struct node{    int ll,rr,v;}t[N*6];int bulid(int x,int y,int idx){    t[idx].ll=x,t[idx].rr=y;    if(x==y) return t[idx].v=l[x];    int mid=(x+y)/2;    return t[idx].v=min(bulid(x,mid,L(idx)),bulid(mid+1,y,R(idx)));}int query(int x,int y,int idx){    if(x==t[idx].ll && y==t[idx].rr) return t[idx].v;    int mid=(t[idx].ll+t[idx].rr)/2;    if(x>mid) return query(x,y,R(idx));    if(y<=mid) return query(x,y,L(idx));    return min(query(x,mid,L(idx)),query(mid+1,y,R(idx)));}int main(){    scanf("%s",bra);    len=strlen(bra);    int tmp=0;    for(int i=1;i<=len;i++)    {        r[i]=r[i-1];        if(bra[i-1]=='(') tmp++;        else if(tmp) r[i]++,tmp--;        l[i]=tmp;    }    bulid(1,len,1);    scanf("%d",&m);    while(m--)    {        int a,b;        scanf("%d%d",&a,&b);        if(a==b) {cout<<0<<endl;continue;}        int ans=min(r[b]-r[a-1],r[b]-r[a-1]-l[a-1]+query(a,b,1));        cout<<ans*2<<endl;    }    return 0;}

Codeforces generation C. sereja and brackets [Line Segment tree]