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++11 (judge verdict: Accepted)

#include <vector>
#include <iostream>

using namespace std;

const uint INF = (uint)(1<<31) - 1;

class RMQ {
  int N;
  vector<uint> dat;

  uint query(int a, int b, int k, int l, int r) {
    if (r <= a || b <= l)
      return INF;
    if (a <= l && r <= b)
      return dat[k];
    uint vl = this->query(a, b, k * 2 + 1, l, (l + r) / 2);
    uint vr = this->query(a, b, k * 2 + 2, (l + r) / 2, r);
    return min(vl, vr);
  }

public:
  RMQ(int n) {
    N = 1;
    while(N < n)
      N *= 2;
    dat.clear();
    dat.resize(2 * N - 1, INF);
  }

  void update(int i, uint x) {
    i += N - 1;
    dat[i] = x;
    while (i > 0) {
      i = (i - 1) / 2;
      dat[i] = min(dat[i * 2 + 1], dat[i * 2 + 2]);
    }
  }

  uint query(int a, int b) {
    return query(a, b, 0, 0, N);
  }
};

int main(void) {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int n, q;
  cin >> n >> q;
  RMQ r(n);
  for (int i = 0; i < q; i++) {
    int c, x, y;
    cin >> c >> x >> y;
    if (c == 0)
      r.update(x, y);
    else
      cout << r.query(x, ++y) << endl;
  }
}