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++11 (judge verdict: Accepted)

#include <bits/stdc++.h>
#define REP(i,n) for(int i=0; i<(int)(n); ++i)

using namespace std;

struct SegmentTree{
    int n;
    vector<int> data;
    SegmentTree(int N){
        n = 1;
        while(n < N) n *= 2;
        data.assign(2 * n, INT_MAX);
    }
    inline void update(int k, int x){
        k += n - 1;
        data[k] = x;
        while(k > 0){
            k = (k - 1) / 2;
            data[k] = min(data[2 * k + 1], data[2 * k + 2]);
        }
    }
    inline int query(int a, int b){
        return query(a, b, 0, 0, n);
    }
    inline int query(int a, int b, int k, int l, int r){
        if(b <= l || r <= a){
            return INT_MAX;
        }else if(a <= l && r <= b){
            return data[k];
        }else{
            int vl = query(a, b, 2 * k + 1, l, (l + r) / 2);
            int vr = query(a, b, 2 * k + 2, (l + r) / 2, r);
            return min(vl, vr);
        }
    }
    void print(){
        for(int i = 0; i < 2 * n - 1; i++){
            printf("data[%d] = %d\n", i, data[i]);
        }
    }
};

int main(){
    int N, Q;
    while(cin >> N >> Q){
        SegmentTree tree(N);
        REP(i, Q){
            int com;
            cin >> com;
            if(com == 0){
                int k, x;
                cin >> k >> x;
                tree.update(k, x);
            }else{
                int a, b;
                cin >> a >> b;
                cout << tree.query(a, b + 1) << endl;
            }
        }
    }
    return 0;
}