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 <iostream>
#include <algorithm>
#define INF 2147483647
using namespace std;
int tree[(1<<18)];
int n,q;

void update(int i,int x) {
  i+=(n-1);
  tree[i] = x;
  while(i!=0){
    i=(i-1)/2;
    tree[i] = min(tree[i*2+1],tree[i*2+2]);
  }
}

int find(int a ,int b,int ima,int L,int R){
  if(R<=a || b<=L) return INF;
  if(a<=L && R<=b) return tree[ima];
  int M = (R+L)/2;
  return min(find(a,b,ima*2+1,L,M),find(a,b,ima*2+2,M,R));
}

int main(){
  cin >> n >>q;
  int N=1;
  while(N<n)N*=2;
  n=N;

  for(int i=0;i<n;i++)update(i,INF);
  while(q--){
    int com,x,y;
    cin >>com >> x >> y;
    if(com == 0) update(x,y);
    if(com == 1) cout << find(x,y+1,0,0,n) <<endl;
  }
  return 0;
}