AOJ DSL_2_A Range Minimum Query (RMQ)

Range Minimum Query (RMQ)

Write a program which manipulates a sequence A = {a₀,a₁,...,a_n−1} with the following operations:

• find(s,t): report the mimimum element in a_s,a_s+1,...,a_t.
• update(i,x): change a_i to x.

Note that the initial values of a_i (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(x_i,y_i) and '1' denotes find(x_i,y_i).

Output

For each find operation, print the minimum element.

Constraints

• 1≤n≤100000
• 1≤q≤100000
• If com_i is 0, then 0≤x_i<n, 0≤y_i<2³¹−1.
• If com_i is 1, then 0≤x_i<n, 0≤y_i<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

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带修改的区间最小值查询，线段树模板题。压了压常数，又打榜了。

 #include <cstdio>

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

 #define siz 10000000

 char buf[siz], *bit = buf;

 inline int nextInt(void) {
     register int ret = ;
     register int neg = false;

     for (; *bit < ''; ++bit)
         if (*bit == '-')neg ^= true;

     for (; *bit >= ''; ++bit)
         ret = ret *  + *bit - '';

     return neg ? -ret : ret;
 }

 #define inf 2147483647

 int n, m, mini[];

 int find(int t, int l, int r, int x, int y) {
     if (x <= l && r <= y)
         return mini[t];
     int mid = (l + r) >> ;
     if (y <= mid)
         return find(t << , l, mid, x, y);
     if (x > mid)
         return find(t <<  | , mid + , r, x, y);
     else
         return min(
             find(t << , l, mid, x, mid),
             find(t <<  | , mid + , r, mid + , y)
         );
 }

 void update(int t, int l, int r, int x, int y) {
     if (l == r)mini[t] = y;
     else {
         int mid = (l + r) >> ;
         if (x <= mid)
             update(t << , l, mid, x, y);
         else
             update(t <<  | , mid + , r, x, y);
         mini[t] = min(mini[t << ], mini[t <<  | ]);
     }
 }

 signed main(void) {
     fread(buf, , siz, stdin);

     n = nextInt();
     m = nextInt();

     for (int i = ; i <= (n << ); ++i)mini[i] = inf;

     for (int i = ; i <= m; ++i) {
         int c = nextInt();
         int x = nextInt();
         int y = nextInt();
         if (c)    // find(x, y)
             printf("%d\n", find(, , n, x + , y + ));
         else    // update(x, y)
             update(, , n, x + , y);
     }

 //    system("pause");
 }

@Author: YouSiki