HDU 5443 The Water Problem (ST算法)

题目链接：HDU 5443

Problem Description

In Land waterless, water is a very limited resource. People always fight for the biggest source of water. Given a sequence of water sources with (a_1,a_2,a_3,...,a_n) representing the size of the water source. Given a set of queries each containing (2) integers (l) and (r), please find out the biggest water source between (a_l) and (a_r).

Input

First you are given an integer (T(Tle 10)) indicating the number of test cases. For each test case, there is a number (n(0le nle 1000)) on a line representing the number of water sources. (n) integers follow, respectively (a_1,a_2,a_3,...,a_n), and each integer is in ({1,...,10^6}). On the next line, there is a number (q(0le qle 1000)) representing the number of queries. After that, there will be (q) lines with two integers (l) and (r(1le lle rle n)) indicating the range of which you should find out the biggest water source.

Output

For each query, output an integer representing the size of the biggest water source.

Sample Input

3
1
100
1
1 1
5
1 2 3 4 5
5
1 2
1 3
2 4
3 4
3 5
3
1 999999 1
4
1 1
1 2
2 3
3 3

Sample Output

100
2
3
4
4
5
1
999999
999999
1

Source

2015 ACM/ICPC Asia Regional Changchun Online

Solution

题意

给定 (n) 个数，(q) 个询问，每个询问包含 (l) 和 (r)，求区间 ([l, r]) 内的最大值。

思路

ST算法

(RMQ) 问题。ST 算法模板题。预处理时间 (O(nlogn))，查询时间 (O(1))。

Code

#include <bits/stdc++.h>
using namespace std;
const int maxn = 1010;

int a[maxn];
int f[maxn][11];
int n;

void st_prework() {
    for(int i = 1; i <= n; ++i) f[i][0] = a[i];
    int t = log(n) / log(2) + 1;
    for(int j = 1; j < t; ++j) {
        for(int i = 1; i <= n - (1 << j) + 1; ++i) {
            f[i][j] = max(f[i][j - 1], f[i + (1 << (j - 1))][j - 1]);
        }
    }
}

int st_query(int l, int r) {
    int k = log(r - l + 1) / log(2);
    return max(f[l][k], f[r - (1 << k) + 1][k]);
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    int T;
    cin >> T;
    while(T--) {
        cin >> n;
        for(int i = 1; i <= n; ++i) {
            cin >> a[i];
        }
        st_prework();
        int q;
        cin >> q;
        for(int i = 0; i < q; ++i) {
            int l, r;
            cin >> l >> r;
            cout << st_query(l, r) << endl;
        }
    }
    return 0;
}