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 <vector>

using namespace std;

#define rep(i,x) for(int i=0;i<(int)(x);++i)

const long long inf = (1ll<<31) - 1;

struct SegTree { // ??????min,max
	int n;
	vector<int> Min;
	void init(int n_) {
		n = 1;
		while (n < n_) n *= 2;
		Min.resize(2 * n);
		for (int i = 0; i < 2 * n - 1; ++i) {
			Min[i] = inf;
		}
	}
	void update(int idx, int val) {
		idx += n - 1;
		Min[idx] = val;
		while (idx > 0) {
			idx = (idx - 1) / 2;
			Min[idx] = min(Min[idx * 2 + 1], Min[idx * 2 + 2]);
		}
	}
	int get_min(int a, int b, int k = 0, int l = 0, int r = -1) { // [a, b)???min
		if (r == -1) r = n;
		if (r <= a || b <= l) return inf;
		if (a <= l && r <= b) return Min[k];
		int vl = get_min(a, b, k * 2 + 1, l, (l + r) / 2);
		int vr = get_min(a, b, k * 2 + 2, (l + r) / 2, r);
		return min(vl, vr);
	}
};

signed main()
{
	int n, q; scanf("%d%d", &n, &q);

	SegTree seg;

	seg.init(n);

	rep(i, q) {
		int type, x, y;
		scanf("%d%d%d", &type, &x, &y);

		if (type % 2 == 0) {
			seg.update(x, y);
		} else {
			printf("%d\n", seg.get_min(x, y + 1));
		}
	}
}