CSU 1809 Parenthesis

Description

1 p

2…p

n

ai and p

bi

Parenthesis sequence S is balanced if and only if:

S is empty;

or there exists 

balanced parenthesis sequence A,B such that S=AB;

or there exists 

balanced parenthesis sequence S' such that S=(S').

Input

The input contains at most 30 sets. For each set:

5,1≤q≤10

5).

1 p

2…p

n.

i,b

i (1≤a

i,b

i≤n,a

i≠b

i).

Output

Sample Input

4 2
(())
1 3
2 3
2 1
()
1 2      

Sample Output

#include<set>
#include<map>
#include<ctime>
#include<cmath>
#include<stack>
#include<queue>
#include<bitset>
#include<cstdio>
#include<string>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
#define rep(i,j,k) for (int i = j; i <= k; i++)
#define per(i,j,k) for (int i = j; i >= k; i--)
#define loop(i,j,k) for (int i = j;i != -1; i = k[i])
#define lson x << 1, l, mid
#define rson x << 1 | 1, mid + 1, r
#define ff first
#define ss second
#define mp(i,j) make_pair(i,j)
#define pb push_back
#define pii pair<int,LL>
#define in(x) scanf("%d", &x);
using namespace std;
typedef long long LL;
const int low(int x) { return x&-x; }
const double eps = 1e-9;
const int INF = 0x7FFFFFFF;
const int mod = 1e9 + 7;
const int N = 1e5 + 10;
int T, n, m, l, r;
char s[N];
int f[N][20], lg[N];

int rmq(int l, int r)
{
  int k = lg[r - l + 1];
  return min(f[l][k], f[r - (1 << k) + 1][k]);
}

int main()
{
  while (scanf("%d%d", &n, &m) != EOF)
  {
    scanf("%s", s + 1);
    int x = 0;  lg[0] = -1;
    rep(i, 1, n)
    {
      x += s[i] == '(' ? 1 : -1;
      f[i][0] = x;    lg[i] = lg[i / 2] + 1;
    }
    for (int j = 1; (1 << j) <= n; j++)
    {
      rep(i, 1, n)
      {
        if (i + (1 << j - 1) > n) break;
        f[i][j] = min(f[i][j - 1], f[i + (1 << j - 1)][j - 1]);
      }
    }
    while (m--)
    {
      scanf("%d%d", &l, &r);
      if (l > r) swap(l, r);
      if (s[l] == s[r] || s[l] == ')') { printf("Yes\n"); continue; }
      if (rmq(l, r - 1) - 2 < 0) printf("No\n"); else printf("Yes\n");
    }
  }
  return 0;
}