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 $a_1,a_2,a_3,\dots ,a_n$ 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 $T(T\le 10)$ indicating the number of test cases. For each test case, there is a number $n(0\le n\le 1000)$ on a line representing the number of water sources. $n$ integers follow, respectively $a_1,a_2,a_3,\dots ,a_n$, and each integer is in $1,\dots ,{10}^6$. On the next line, there is a number $q(0\le q\le 1000)$ representing the number of queries. After that, there will be $q$ lines with two integers $l$ and $r(1\le l\le r\le n)$ 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;
}