POJ 3264 Balanced Lineup

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这是一道关于计算不同连续区间内奶牛高度差的问题。农夫约翰有N头奶牛，每天按固定顺序排队挤奶。他打算挑选一部分奶牛玩终极飞盘游戏，要求选中的奶牛身高差异不能太大。输入包含奶牛的数量、查询次数及每个奶牛的身高，接着是多个查询区间。任务是为每个查询区间计算最矮和最高奶牛之间的高度差。

Balanced Lineup

https://vjudge.net/problem/POJ-3264
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
C++

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int MAXN = 50000 + 10;
const int INF = 0x3f3f3f3f;

int num[MAXN];
int MAX, MIN;

struct node{
	int l, r, maxn, minn;
}segTree[MAXN << 2];

void build(int x, int l, int r)
{
	segTree[x].l = l;
	segTree[x].r = r;
	if(l == r)
	{
		segTree[x].maxn = num[l];
		segTree[x].minn = num[l];
		return;
	}
	int mid = l + r >> 1;
	build(x << 1, l, mid);
	build(x << 1 | 1, mid + 1, r);
	segTree[x].maxn = max(segTree[x << 1].maxn, segTree[x << 1 | 1].maxn);
	segTree[x].minn = min(segTree[x << 1].minn, segTree[x << 1 | 1].minn);
}

void query(int x, int l, int r)
{
	if(l <= segTree[x].l && segTree[x].r <= r)
	{
		MAX = max(MAX, segTree[x].maxn);
		MIN = min(MIN, segTree[x].minn);
		return;
	}
	int mid = segTree[x].l + segTree[x].r >> 1;
	if(r <= mid) query(x << 1, l, r);
	else if(l > mid) query(x << 1 | 1, l, r);
	else{
		query(x << 1, l, mid);
		query(x << 1 | 1, mid + 1, r);
	}
}

int main()
{
	int n, m;
	scanf("%d%d", &n, &m);
	for(int i = 1; i <= n; i ++)
		scanf("%d", &num[i]);
	build(1, 1, n);
	int a, b;
	for(int i = 1; i <= m; i ++)
	{
		scanf("%d%d", &a, &b);
		MAX = -INF, MIN = INF;
		query(1, a, b);
		printf("%d\n", MAX - MIN);
	}
	return 0;
}