POJ 3264 Balanced Lineup（最大最小差值线段树水题）

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本文深入探讨了游戏开发领域的关键技术，包括游戏引擎、编程语言、硬件优化等，并重点介绍了AI音视频处理在游戏中的应用，如语音识别、图像处理、AR特效等。同时，文章还涉及了自动化测试、性能优化等现代开发流程，旨在为游戏开发者提供全面的技术指导。

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Balanced Lineup

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 37122		Accepted: 17383
Case Time Limit: 2000MS

Description

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

Sample Output

6
3
0

Source

USACO 2007 January Silver

最大最小差值

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<set>
#include<map>


#define L(x) (x<<1)
#define R(x) (x<<1|1)
#define MID(x,y) ((x+y)>>1)

#define eps 1e-8
typedef __int64 ll;

#define fre(i,a,b)  for(i = a; i <b; i++)
#define free(i,b,a) for(i = b; i >= a;i--)
#define mem(t, v)   memset ((t) , v, sizeof(t))
#define ssf(n)      scanf("%s", n)
#define sf(n)       scanf("%d", &n)
#define sff(a,b)    scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define pf          printf
#define bug         pf("Hi\n")

using namespace std;

#define INF 0x3f3f3f3f
#define N 50005

struct stud{
int le,ri;
int mi,ma;
}f[N*4];

int a[N],mi,ma;

void build(int pos,int le,int ri)
{
	f[pos].le=le;
	f[pos].ri=ri;
	if(le==ri)
	{
		f[pos].mi=f[pos].ma=a[le];
		return ;
	}

    int mid=MID(le,ri);
    build(L(pos),le,mid);
    build(R(pos),mid+1,ri);

    f[pos].ma=max(f[L(pos)].ma,f[R(pos)].ma);
    f[pos].mi=min(f[L(pos)].mi,f[R(pos)].mi);
}

void query(int pos,int le,int ri)
{
	if(f[pos].le==le&&f[pos].ri==ri)
	{
		mi=min(mi,f[pos].mi);
		ma=max(ma,f[pos].ma);
		return ;
	}

   int mid=MID(f[pos].le,f[pos].ri);

   if(mid>=ri)
	query(L(pos),le,ri);
   else
	if(mid<le)
	 query(R(pos),le,ri);
   else
   {
   	 query(L(pos),le,mid);
   	 query(R(pos),mid+1,ri);
   }

}

int main()
{
	int i,j;
	int n,m;
	while(~sff(n,m))
	{
		fre(i,1,n+1)
		 sf(a[i]);

		 build(1,1,n);
		 int le,ri;
		 while(m--)
		 {
		 	sff(le,ri);
		 	mi=INF;
		 	ma=-1;
            query(1,le,ri);
		 	pf("%d\n",ma-mi);
		 }
	}
   return 0;
}