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<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<climits>
#include<iostream>
#include<sstream>
#include<utility>
#include<map>
#include<vector>
#include<queue>
#include<algorithm>
#include<set>
using namespace std;
typedef long long ll;
typedef pair<int,int>P;
const int MAX_N=100009;
int tree[3*MAX_N];
int inv=INT_MAX;

int N,Q;

void init(int _N,int iniv)
{
	for(N=1;N<=_N;)
		N*=2;
	for(int i=0;i<2*N-1;i++)
		tree[i]=iniv;
}

void update(int p,int a)
{
	p+=N-1;
	tree[p]=a;
	for(;p>0;)
	{
		p=(p-1)/2;
		tree[p]=min(tree[p*2+1],tree[p*2+2]);
	}
}

int query(int l,int r,int k,int nl,int nr)
{
	if(nr<=l||r<=nl)return inv;
	if(l<=nl&&nr<=r)return tree[k];
	 
	int vl=query(l,r,k*2+1,nl,(nl+nr)/2);
	int vr=query(l,r,k*2+2,(nl+nr)/2,nr);
	return min(vl,vr);
}


int main()
{
	scanf("%d%d",&N,&Q);
	init(N,inv);
	for(int i=0;i<Q;i++)
	{
		int con,x,y;
		scanf("%d%d%d",&con,&x,&y);
		if(!con)
			update(x,y);
		else
			printf("%d\n",query(x,y+1,0,0,N));
	}
}