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>
#include <stdlib.h>

#define INF 2147483647
#define RMQ_SIZE 262144  // 2^18

int tree[RMQ_SIZE];

int dfs(int a, int b, int k, int l, int r);
int find(int first, int second);
void update(int pos, int value);
int min(int a, int b);


int main(void)
{
    int n, q, com, x, y, i;

    scanf("%d %d", &n ,&q);

    for (i = 0; i < n; i++) {
        update(i, INF);
    }

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

/*
 * a??????b???????????????????????¢?´¢????????????????±????.
 * ?????¨???tree[k]???????????????.
 * tree[k]???l??????r?????§???????°?????????????.
 */
int dfs(int a,int b,int k,int l,int r)
{
    int m;
    //* a~b??¨l~r??????????????????????????¨????????¨?????????.
    if (r <= a || b <= l) {
        return INF;
    }
    //* tree[k]?????¬?????????????±?????????\???a~b?????????????????¨???.
    if (a <= l && r <= b) {
        return tree[k];
    }
    m = (l+r)/2;
    return min(dfs(a, b, k*2, l, m), dfs(a, b, k*2+1, m, r));
}

/*
 * ??????????????????[right,left)??§???????????¨?????¨???
 */
int find(int right, int left)
{
    return dfs(right, left, 1, 0, RMQ_SIZE/2);
}

/*
 * pos?????????????????????value????????´??????.
 * ??????????????¨??¨?????????????????????.
 */
void update(int pos, int value)
{
    pos += RMQ_SIZE/2;
    tree[pos] = value;
    /*
     * ?????????????????§????°?????????´??°????????????.
     * ???????????????????????????.
     * ????????´????????´??°???????????????????????£???????????§return??????.
     */
    while (pos) {
        pos = pos/2;
        tree[pos] = min(tree[pos*2], tree[pos*2+1]);
    }
}

int min(int a, int b)
{
    return a < b ? a: b;
}