HDU 5443 The Water Problem (ST算法)

題目連結：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 \(T(T\le 10)\) 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 \(a_1,a_2,a_3,...,a_n\), and each integer is in \({1,...,10^6}\). 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

題意

給定 \(n\) 個數，\(q\) 個詢問，每個詢問包含 \(l\) 和 \(r\)，求區間 \([l, r]\) 内的最大值。

思路

ST算法

\(RMQ\) 問題。ST 算法模闆題。預處理時間 \(O(nlogn)\)，查詢時間 \(O(1)\)。

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;
}