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: Runtime Error)

#include <stdio.h>          // printf(), scanf()

#define MAX_N 100000
#define MAX_INT 0x7fffffff
#define min(a, b) ((a) < (b) ? (a) : (b))

int dat[2 * MAX_N - 1];
int n;

void
init_rmq(int m)
{
	int i;

	n = 1;
	while (n < m)
		n *= 2;

	for (i = 0; i < 2 * n - 1; ++i)
		dat[i] = MAX_INT;
}

void
update(int k, int a)
{
	k += n - 1;
	dat[k] = a;
	while (k > 0)
	{
		k = (k - 1) / 2;
		dat[k] = min(dat[k * 2 + 1], dat[k * 2 + 2]);
	}
}

int
query(int a, int b, int k, int l, int r)
{
	if (r <= a || b <= l)
		return MAX_INT;

	if (a <= l && r <= b)
		return dat[k];

	int vl = query(a, b, k * 2 + 1, l, (l + r) / 2);
	int vr = query(a, b, k * 2 + 2, (l + r) / 2, r);
	return min(vl, vr);
}

int
main(int argc, char** argv)
{
	int q;
	int c, x, y;
	int i;

	scanf("%d %d", &n, &q);
	init_rmq(n);
	for (i = 0; i < q; ++i)
	{
		scanf("%d %d %d", &c, &x, &y);
		if (c == 0)
			update(x, y);
		else
			printf("%d\n", query(x, y + 1, 0, 0, n));
	}

	return 0;
}