poj3264 Balanced Lineup（求区间的最大值与最小值之差）

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本博客探讨了如何通过编程解决农民约翰组织奶牛玩飞盘游戏时，确保奶牛队伍高度差异不大的问题。包括输入解析、构建数据结构、实现区间查询算法等步骤，最终确定每组奶牛中高度差异最小的解决方案。

Balanced Lineup

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 37869		Accepted: 17751
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

解题思路：先求最大值，再求最小值

/*******************************************************************************
 * Author :          jinbao
 * Email :           dongjinbao913106840144@gmail.com
 * Last modified :   2015-05-11 20:30
 * Filename :        poj3264.cpp
 * Description :     
 * *****************************************************************************/
#include <iostream>
#include <stdio.h>
using namespace std;
#define min(x,y) x<y?x:y
#define max(x,y) x>y?x:y
const int MAX = 50000+1;
struct cow{
	int Min,Max;
}p[4*MAX];
int a[MAX];
void build(int node,int begin,int end){
	if (begin==end){
		scanf("%d",&a[begin]);
		p[node].Min=p[node].Max=a[begin];
	}
	else{
		build(node*2,begin,(begin+end)/2);
		build(node*2+1,(begin+end)/2+1,end);
		p[node].Min=min(p[node*2].Min,p[node*2+1].Min);
		p[node].Max=max(p[node*2].Max,p[node*2+1].Max);
	}
}
int query(int node,int begin,int end,int left,int right,int flag){
	if (end<left || begin>right){
		if (flag==0)
			return 0xffffff;
		return -1;
	}
	if (begin>=left && end<=right){
		if (flag==0)
			return p[node].Min;
		return p[node].Max;
	}
	int m=query(2*node,begin,(begin+end)/2,left,right,flag);
	int n=query(2*node+1,(begin+end)/2+1,end,left,right,flag);
	if (flag==0)
		return min(m,n);
	return max(m,n);
}
int main(){
	int n,q,l,r,Min,Max;
	while (~scanf("%d%d",&n,&q)){
		build(1,1,n);
		while (q--){
			scanf("%d%d",&l,&r);
			int ans = query(1,1,n,l,r,1) - query(1,1,n,l,r,0);
			printf("%d\n",ans);
		}
	}

	return 0;
}