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>
int M=(1<<31)-1;
int MIN(int a,int b){return a<b?a:b;}
int main(){
  int n,m,B=200,i,q,x,y,a[100010],b[1010],min;
  scanf("%d %d",&n,&m);
  for(i=0;i<100010;i++)a[i]=M;
  for(i=0;i<1010;i++)b[i]=M;
  while(m--){
    scanf("%d %d %d",&q,&x,&y);
    if(q){
      min=M;
      while(x<y+1){
	if(x%B==0&&x+B<=y+1){
	  min=MIN(min,b[x/B]);
	  x+=B;
	}
	else min=MIN(min,a[x++]);
      }
      printf("%d\n",min);
    }
    else{
      a[x]=y;
      b[x/B]=M;
      for(i=x-x%B;i<MIN(n,x-x%B+B);i++)b[x/B]=MIN(b[x/B],a[i]);
    }
    //for(i=0;i<n;i++)printf("%d ",a[i]);printf("\n");
    //for(i=0;i<=x/B;i++)printf("%d ",b[i]);printf("\n");
  }
  return 0;
}