【线段树】Balanced Lineup(区间最值)

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

题意: 给N个值，Q个查询，每次查询[l,r]中的最大值减最小值。
其实就是一个线段树求区间最值的模板题。O(n)的建树，O(lgn)的查询。这题会卡输入的时间，用scanf提高输入效率。同时这道题也可以剪枝。下面直接上代码。

#include <iostream>
#define MAX 50010
using namespace std;

int A[MAX];
int N, Q;

struct Node{
	int Max, Min; 
	Node(){Max = 0; Min = INT_MAX;}
}node[MAX<<2];

inline void PushUp(int rt)
{
	node[rt].Max = max(node[rt<<1].Max, node[rt<<1|1].Max);
	node[rt].Min = min(node[rt<<1].Min, node[rt<<1|1].Min);
}

void build(int rt, int l, int r)
{
	if(l == r)
	{
		node[rt].Max = A[l];
		node[rt].Min = A[l];
		return;
	}
	int mid = l+r>>1;
	build(rt<<1, l, mid);
	build(rt<<1|1, mid+1, r);
	PushUp(rt);
}

void Query(int rt, int l, int r, int L, int R, int &ans1, int &ans2)
{
	if(L <= l && R >= r)
	{
		ans1 = node[rt].Max;
		ans2 = node[rt].Min;
		return;
	}
	//当前最大值比区间最大值大且当前最小值比区间最小值小，那么这个区间就不会更新ans1,ans2的值，无需递归	 
	if(node[rt].Max <= ans1 && node[rt].Min >= ans2) return;	
	int mid = l+r>>1, x1 = 0, x2 = 0, x11 = INT_MAX, x22 = INT_MAX;
	if(L <= mid) Query(rt<<1, l, mid, L, R, x1, x11);
	if(R > mid) Query(rt<<1|1, mid+1, r, L, R, x2, x22);
	ans1 = max(x1, x2);
	ans2 = min(x11, x22);
}

int main()
{
	while(cin>>N>>Q)
	{
		for(int i = 1;i <= N;i++) scanf("%d",&A[i]);
		build(1,1,N);
		while(Q--)
		{
			int L, R, Max = 0, Min = INT_MAX;
			scanf("%d%d",&L, &R);
			Query(1,1,N,L,R,Max,Min);
			cout<<Max-Min<<endl;
		}	
	}
	return 0;
}