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++14 (judge verdict: Accepted)

#include <bits/stdc++.h>
#define int long long int
#define MOD(x) ((x % MOD_N) + MOD_N) % MOD_N
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define FORE(i,a,b) for(int i=(a);i<=(b);++i)
#define RFOR(i,a,b) for(int i=(b)-1;i>=(a);--i)
#define RFORE(i,a,b) for(int i=(b);i>=(a);--i)
#define REP(i,n) FOR(i,0,n)
#define RREP(i,n) RFOR(i,0,n)
#define ALL(c) (c).begin(),(c).end()
#define RALL(c) (c).rbegin(),(c).rend()
#define SORT(c) sort(ALL(c))
#define RSORT(c) sort(RALL(c))
#define SZ(c) (int)((c).size())
#define EACH(i,v) for(auto i=v.begin();i!=v.end();++i)
#define REACH(i,v) for(auto i=v.rbegin();i!=v.rend();++i)
#define LB(c,x) distance((c).begin(),lower_bound(ALL(c),x))
#define UB(c,x) distance((c).begin(),upper_bound(ALL(c),x))
#define COUNT(c,x) (lower_bound(ALL(c),x)-upper_bound(ALL(c),x))
#define UNIQUE(c) SORT(c); (c).erase(unique(ALL(c)),(c).end());
#define COPY(c1,c2) copy(ALL(c1),(c2).begin())
#define EXIST(s,e) (bool)((s).find(e)!=(s).end())
#define PB push_back
#define MP make_pair
#define DUMP(x)  cerr<<#x<<" = "<<(x)<<endl;
#define NL cerr<<endl;
using namespace std;
template<typename T,typename U> using P=pair<T,U>;
template<typename T> using V=vector<T>;
template<typename T>bool chmax(T&a,T b){if(a<b){a=b;return true;}return false;}
template<typename T>bool chmin(T&a,T b){if(a>b){a=b;return true;}return false;}
template<typename T>T sum(vector<T>&v){return accumulate(ALL(v),T());}
template<typename T>T sum(vector<T>&v,int a,int b){return accumulate(v.begin()+a,v.begin()+b,T());}
template<typename T>T max(vector<T>&v){return *max_element(ALL(v));}
template<typename T>T min(vector<T>&v){return *max_element(ALL(v));}
template<typename T>T max_index(vector<T>&v){return distance((v).begin(),max_element(ALL(v)));}
template<typename T>T min_index(vector<T>&v){return distance((v).begin(),min_element(ALL(v)));}

struct edge { int to, cost; };

template<typename T>auto&operator<<(ostream&s,const vector<T>&v){s<<"[";bool a=1;for(auto e:v){s<<(a?"":" ")<<e;a=0;}s<<"]";return s;}
template<typename T,typename U>auto&operator<<(ostream&s,const pair<T,U>&p){s<<"("<<p.first<<","<<p.second<<")";return s;}
template<typename T>auto&operator<<(ostream&s,const set<T>&st){s<<"{";bool a=1;for(auto e:st){s<<(a?"":" ")<<e;a=0;}s<<"}";return s;}
template<typename T,typename U>auto&operator<<(ostream&s,const map<T,U>&m){s<<"{";bool a=1;for(auto e:m){s<<(a?"":" ")<<e.first<<":"<<e.second;a=0;}s<<"}";return s;}

const int INF = 1e18;
const int MOD_N = 1e9+7;

template<typename T> class SegmentTree
{
   private:
      int n;
      vector<T> node;
   public:
      SegmentTree(vector<T> v) {
         int sz = v.size();
         n = 1; while(n < sz) n *= 2;
         node.resize(2*n-1);

         for(int i=0; i<sz; i++) node[i+n-1] = v[i];
         for(int i=n-2; i>=0; i--) node[i] = min(node[2*i+1], node[2*i+2]);
      }
      void update(int x, T val) {
         x += (n - 1);
         node[x] = val;
         while(x > 0) {
            x = (x - 1) / 2;
            node[x] = min(node[2*x+1], node[2*x+2]);
         }
      }
      int find(int a, int b, int k=0, int l=0, int r=-1) {
         // return min in [a, b)
         if(r < 0) r = n;
         if(r <= a || b <= l) return INF;
         if(a <= l && r <= b) return node[k];
         int vl = find(a, b, 2*k+1, l, (l+r)/2);
         int vr = find(a, b, 2*k+2, (l+r)/2, r);
         return min(vl, vr);
      }
};

signed main()
{
   int n, q; cin >> n >> q;
   V<int> c(q), x(q), y(q);
   REP(i, q) {
      cin >> c[i] >> x[i] >> y[i];
   }

   V<int> v(n, (1LL<<31)-1);
   SegmentTree<int> st(v);
   REP(i, q) {
      if (c[i] == 0) {
         st.update(x[i], y[i]);
      } else {
         cout << st.find(x[i], y[i]+1) << endl;
      }
   }

   return 0;
}