POJ 3264-Balanced Lineup（段树：单点更新，间隔查询）

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.

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cctype>
#include <cmath>
#include <cstdlib>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <list>
#define LL long long
using namespace std;
const int INF=1<<27;
const int maxn=200010;
LL minn[maxn],maxx[maxn];
void update(LL root,LL l,LL r,LL p,LL v)//单点更新
{
	if(l==r) maxx[root]=v;minn[root]=v;
	if(l<r)
	{
		LL mid=(l+r)/2;
		if(p<=mid) update(root*2,l,mid,p,v);
		else update(root*2+1,mid+1,r,p,v);
		maxx[root]=max(maxx[root*2],maxx[root*2+1]);
		minn[root]=min(minn[root*2],minn[root*2+1]);
	}
}
LL query_min(LL root,LL l,LL r,LL ql,LL qr)
{
	LL mid=(l+r)/2,ans=INF;
	if(ql<=l&&qr>=r) return minn[root];
	if(ql<=mid) ans=min(ans,query_min(root*2,l,mid,ql,qr));
	if(qr>mid) ans=min(ans,query_min(root*2+1,mid+1,r,ql,qr));
	return ans;
}
LL query_max(LL root,LL l,LL r,LL ql,LL qr)
{
	LL mid=(l+r)/2,ans=-INF;
	if(ql<=l&&qr>=r) return maxx[root];
	if(ql<=mid) ans=max(ans,query_max(root*2,l,mid,ql,qr));
	if(qr>mid) ans=max(ans,query_max(root*2+1,mid+1,r,ql,qr));
	return ans;
}
int main()
{
  int N,Q,i,v;
  while(~scanf("%lld%lld",&N,&Q))
  {
  	for(i=1;i<=N;i++)
	{
		scanf("%lld",&v);
		update(1,1,N,i,v);
	}
	while(Q--)
	{
		int ql,qr;
		scanf("%lld%lld",&ql,&qr);
		printf("%lld\n",query_max(1,1,N,ql,qr)-query_min(1,1,N,ql,qr));
	}
  }
  return 0;
}