Balanced Lineup——线段树查询-优快云博客

本文介绍了一种利用线段树进行区间最大值和最小值查询的算法，适用于解决涉及大量区间查询的问题，如在固定范围内寻找高度差最大的奶牛组。通过构建线段树，可以高效地获取任意指定区间的最大值和最小值，从而计算出最大高度差。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

For the daily milking, Farmer John’s N cows (1 ≤ N ≤ 50,000) always line up in the same order. One day Farmer John decides to organize a game of Ultimate Frisbee with some of the cows. To keep things simple, he will take a contiguous range of cows from the milking lineup to play the game. However, for all the cows to have fun they should not differ too much in height.

Farmer John has made a list of Q (1 ≤ Q ≤ 200,000) potential groups of cows and their heights (1 ≤ height ≤ 1,000,000). For each group, he wants your help to determine the difference in height between the shortest and the tallest cow in the group.

Input
Line 1: Two space-separated integers, N and Q.
Lines 2…N+1: Line i+1 contains a single integer that is the height of cow i
Lines N+2…N+Q+1: Two integers A and B (1 ≤ A ≤ B ≤ N), representing the range of cows from A to B inclusive.
Output
Lines 1…Q: Each line contains a single integer that is a response to a reply and indicates the difference in height between the tallest and shortest cow in the range.
Sample Input
6 3
1
7
3
4
2
5
1 5
4 6
2 2
Sample Output
6
3
0

虽然只是个题但还是看不透这个农夫

#include<iostream>
#include<algorithm>
#include<cstring>
#include<stdio.h>
#define MAX 200005
using namespace std;
int tree1[MAX * 4], insert[MAX];
int tree2[MAX * 4];
void update(int k)
{
	tree1[k] = max(tree1[2 * k], tree1[2 * k + 1]);
	tree2[k] = min(tree2[2 * k], tree2[2 * k + 1]);
}
void build(int k, int left, int right)// 建树
{
	if (left == right)
	{
		tree1[k] = insert[left];
		tree2[k] = insert[left];
		return;
	}
	int n = (left + right) / 2;
	build(2 * k, left, n);
	build(2 * k + 1, n + 1, right);
	update(k);
}
int query_max(int k, int left, int right, int query_left, int query_right)//查询区间
{
	if (query_left > right || query_right < left)
		return 0;
	if (query_left <= left && query_right >= right)
		return tree1[k];
	int n = (left + right) / 2;
	int res1 = query_max(2 * k, left, n, query_left, query_right);
	int res2 = query_max(2 * k + 1, n + 1, right, query_left, query_right);
	return max(res1, res2);
}
int query_min(int k, int left, int right, int query_left, int query_right)//查询区间
{
	if (query_left > right || query_right < left)
		return 0x3f3f3f;
	if (query_left <= left && query_right >= right)
		return tree2[k];
	int n = (left + right) / 2;
	int res1 = query_min(2 * k, left, n, query_left, query_right);
	int res2 = query_min(2 * k + 1, n + 1, right, query_left, query_right);
	return min(res1, res2);
}
int main()
{
	int n, m;
	while (~scanf("%d%d", &n, &m))
	{
		for (int i = 1; i <= n; i++)
			scanf("%d", &insert[i]);
		build(1, 1, n);
		while (m--)
		{
			int l, r;
			scanf("%d%d", &l, &r);
			int ans = query_max(1, 1, n, l, r) - query_min(1, 1, n, l, r);
			printf("%d\n", ans);
		}
	}
	return 0;
}