PTA The Water Problem (10分)(线段树求区间最值)

The Water Problem (10分)

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 a1 ,a2 ,a3 ,…,an 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 al and ar .

输入格式:

First you are given an integer T (T ≤ 10) indicating the number of test cases. For each test case, there is a number n (0 ≤ n ≤ 1000) on a line representing the number of water sources. n integers follow, respectively a1 ,a2 ,a3 ,…,an , and each integer is in {1,…,10^6 }. On the next line, there is a number q (0 ≤ q ≤ 1000) representing the number of queries. After that, there will be q lines with two integers l and r (1 ≤ l ≤ r ≤ n) indicating the range of which you should find out the biggest water source.

输出格式:

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

输入样例:

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

输出样例:

100
2
3
4
4
5
1
999999
999999
1

解题

该题直接暴力解也行，这里用线段树，二分查找。

代码

#include <algorithm>  //7-5 The Water Problem (10分)
#include <cstring>
#include <iostream>
using namespace std;
const int maxn = 1e3 + 2;
int d[maxn * 4], num[maxn];

void build(int s, int t, int p) {
    
    
    if (s == t) {
    
      //对 [s, t] 区间建立线段树, 当前根的编号为 p
        d[p] = num[s];
        return;
    }
    int m = (s + t) >> 1;
    build(s, m, p * 2);          //最终可以求出d[p * 2]
    build(m + 1, t, p * 2 + 1);  //最终可以求出d[p * 2 + 1]
    d[p] = max(d[p * 2], d[p * 2 + 1]);
}

int getMaxValue(int l, int r, int s, int t, int p) {
    
    
    // 要查询区间 [l ,r] 的最大值
    if (l <= s && r >= t) return d[p];  //所求区间包含了已知区间
    int m = (s + t) / 2;                //二分查找
    if (r <= m) {
    
    
        return getMaxValue(l, r, s, m, p * 2);
    } else if (l > m) {
    
    
        return getMaxValue(l, r, m + 1, t, p * 2 + 1);
    } else {
    
    
        return max(getMaxValue(l, m, s, m, p * 2),
                   getMaxValue(m + 1, r, m + 1, t, p * 2 + 1));
    }
}

int main() {
    
    
    int T, n, q;
    cin >> T;
    while (T--) {
    
    
        // cin >> n;
        scanf("%d", &n);
        for (int j = 1; j <= n; j++) scanf("%d", &num[j]);
        build(1, n, 1);  //建立区间最值线段树

        cin >> q;
        int l, r;
        while (q--) {
    
    
            scanf("%d %d", &l, &r);
            cout << getMaxValue(l, r, 1, n, 1) << endl;  //二分查找
        }
    }

    system("pause");
    return 0;
}

有关线段树：https://blog.csdn.net/qq_45349225/article/details/109338072
详细介绍：https://oi-wiki.org/ds/seg/