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 a[100001];
int com[100001],x[100001],y[100001],b[100001];
int B=2000;
int n,q;
void put(int a[100001]);
int min(int a,int b);
void update(int i,int x);
void find(int s,int t);
int main(void){
  put(a);
  int i;
  scanf("%d %d",&n,&q);
  for(i=0;i<q;i++){
    scanf("%d %d %d",&com[i],&x[i],&y[i]);
    if(com[i]==0)
      update(x[i],y[i]);
    else if(com[i]==1)
      find(x[i],y[i]);
  }
  return 0;
}
int min(int a,int b){
  if(a<=b)
    return a;
  else
    return b;
}
void update(int i,int x){
  int j;
  a[i]=x;
  b[i/B]=2147483647;
  for(j=i-i%B;j<min(i-i%B+B,n);j++){
    b[i/B]=min(b[i/B],a[j]);
  }
}
void find(int s,int t){
  int ans=2147483647;
  int j;
  for(j=s;j<=t;){
    if(j%B==0 && j+B<=t-1){
      ans=min(ans,b[j/B]);
      j+=B;
    }else{
      ans=min(ans,a[j]);
      j++;
    }
  }
  printf("%d\n",ans);
}
void put(int a[100001]){
  int i;
  for(i=0;i<100001;i++)
    a[i]=2147483647,b[i]=2147483647;
}