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 = 0;
    register int neg = false;
    
    for (; *bit < '0'; ++bit)
        if (*bit == '-')neg ^= true;
    
    for (; *bit >= '0'; ++bit)
        ret = ret * 10 + *bit - '0';
    
    return neg ? -ret : ret;
}

#define inf 2147483647

int n, m, mini[400005];

int find(int t, int l, int r, int x, int y) {
    if (x <= l && r <= y)
        return mini[t];
    int mid = (l + r) >> 1;
    if (y <= mid)
        return find(t << 1, l, mid, x, y);
    if (x > mid)
        return find(t << 1 | 1, mid + 1, r, x, y);
    else
        return min(
            find(t << 1, l, mid, x, mid),
            find(t << 1 | 1, mid + 1, r, mid + 1, y)
        );
}

void update(int t, int l, int r, int x, int y) {
    if (l == r)mini[t] = y;
    else {
        int mid = (l + r) >> 1;
        if (x <= mid)
            update(t << 1, l, mid, x, y);
        else
            update(t << 1 | 1, mid + 1, r, x, y);
        mini[t] = min(mini[t << 1], mini[t << 1 | 1]);
    }
}

signed main(void) {
    fread(buf, 1, siz, stdin);

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

    for (int i = 0; i <= (n << 2); ++i)mini[i] = inf;

    for (int i = 1; i <= m; ++i) {
        int c = nextInt();
        int x = nextInt();
        int y = nextInt();
        if (c)    // find(x, y)
            printf("%d
", find(1, 1, n, x + 1, y + 1));
        else    // update(x, y)
            update(1, 1, n, x + 1, y);
    }

//    system("pause");
}

@Author: YouSiki