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)

// AOJ DSL_2_A Range Minimum Query (RMQ)
// 2018.5.5 bal4u

#include <stdio.h>
#include <stdlib.h>

#define INF 0x7fffffff

#if 1
#define gc() getchar_unlocked()
#define pc(c) putchar_unlocked(c)
#else
#define gc() getchar()
#define pc(c) putchar(c)
#endif
int in()
{
	int n = 0, c = gc();
	do n = 10*n + (c & 0xf), c = gc(); while (c >= '0');
	return n;
}

void out(int n)
{
	int i;
	char ob[20];
	char inf[11] = "2147483647";

	if (!n) pc('0');
	else if (n == INF) for (i = 0; i < 10; i++) pc(inf[i]);
	else {
		i = 0; while (n) ob[i++] = n%10 + '0', n/=10;
		while (i--) pc(ob[i]);
	}
	pc('\n');
}

// セグメント木の初期化：配列の確保と初期値による初期化
int seg[1<<18]; int sz;

void segtree(int n, int init_val)
{
	int i;

	sz = 1; while (sz < n) sz <<= 1;
	n = sz << 1;
//	seg = malloc(n << 2);
	for (i = 0; i < n; i++) seg[i] = init_val;
}

// k番目（0-index）の値を v に変更
void update(int k, int v)
{
	int p;

	k += sz - 1;
	seg[k] = v;
	while (k > 0) {
		k = (k-1) >> 1;
		p = (k << 1)+1;
		v = seg[p+1];
		if (v > seg[p]) v = seg[p];
		seg[k] = v;
	}
}

int query2(int a, int b, int k, int l, int r) 
{
	int m, lmin, rmin;

	if (r <= a || b <= l) return INF;
    if (a <= l && r <= b) return seg[k];
	k <<= 1, m = (l+r) >> 1;
    lmin = query2(a, b, k+1, l, m);
    rmin = query2(a, b, k+2, m, r);
	if (lmin > rmin) lmin = rmin;
    return lmin; 
}

// 空間 [a, b)　内の最小値
int range_min_query(int a, int b)
{
	return query2(a, b+1, 0, 0, sz);
}


int main()
{
	int n, q, op, x, y;

	n = in(), q = in();
	segtree(n, INF);

	while (q--) {
		op = in(), x = in(), y = in();
		if (op) out(range_min_query(x, y));
		else update(x, y);
	}
	return 0;
}