Problem:
Range Minimum Query (RMQ)

Write a program which manipulates a sequence A = {a0, a1, . . . , an-1} with the following operations:

find(s, t): report the minimum element in as, as+1, . . . ,at.

update(i, x): change ai to x.

Note that the initial values of ai (i = 0, 1, . . . , n−1) are 231-1.

Input

n q
com0 x0 y0
com1 x1 y1
...
comq−1 xq−1 yq−1

In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, q queries are given where com represents the type of queries. '0' denotes update(xi, yi) and '1' denotes find(xi, yi).

Output

For each find operation, print the minimum element.

Constraints

1 ≤ n ≤ 100000

1 ≤ q ≤ 100000

If comi is 0, then 0 ≤ xi < n, 0 ≤ yi < 231-1.

If comi is 1, then 0 ≤ xi < n, 0 ≤ yi < n.

Sample Input 1

3 5
0 0 1
0 1 2
0 2 3
1 0 2
1 1 2

Sample Output 1

1
2

Sample Input 2

1 3
1 0 0
0 0 5
1 0 0

Sample Output 2

2147483647
5

Solution in C (judge verdict: Accepted)

#include<stdio.h>
#define ST_SIZE ((1 << 18) - 1)
#define INF 0x7fffffff
#define min(a, b) ((a) < (b) ? (a) : (b))
typedef long long int LL;
LL dat[ST_SIZE];
LL input(void);
void update(LL k, LL a, LL n);
LL minimize(LL a, LL b, LL k, LL l, LL r);
int main (void) {
  int command;
  LL i, n, q, s, t, n2 = 1;
  n = input();
  q = input();
  while (n2 < n) n2 *= 2;
  n = n2;
  for (i = 0; i < 2 * n - 1; i++) dat[i] = INF;
  for (i = 0; i < q; i++) {
    command = input();
    if (!command) {
      s = input();
      t = input();
      update(s, t, n);
    } else {
      s = input();
      t = input();
      printf("%lld\n", minimize(s, t + 1, 0, 0, n));
    }
  }
}

LL input(void)
{
  LL n = 0;
  int c = getchar_unlocked();
  if (c == '-') {
    c = getchar_unlocked();
    do {
      n = 10 * n + c - '0';
      c = getchar_unlocked();
    } while (c >= '0' && c <= '9');
    return -n;
  } else {
    do {
      n = 10 * n + c - '0';
      c = getchar_unlocked();
    } while (c >= '0' && c <= '9');
    return n;
  }
}

void update(LL k, LL a, LL n)
{
  k += n - 1;
  dat[k] = a;
  while (k > 0) {
    k = (k - 1)/ 2;
    dat[k] = min(dat[k * 2 + 1], dat[k * 2 + 2]);
  }
}

LL minimize(LL a, LL b, LL k, LL l, LL r)
{
  if (a >= r || b <= l) return INF;
  else if (a <= l && b >= r) return dat[k];
  else {
    LL vl , vr;
    vl = minimize(a, b, k * 2 + 1, l, (l + r) / 2);
    vr = minimize(a, b, k * 2 + 2 , (l + r) / 2, r);
    return min(vl, vr);
  }
}