poj 3264

Balanced Lineup

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 44506		Accepted: 20902
Case Time Limit: 2000MS

Description

For the dailymilking, Farmer John's N cows (1 ≤ N ≤ 50,000)always line up in the same order. One day Farmer John decides to organize agame of Ultimate Frisbee with some of the cows. To keep things simple, he willtake a contiguous(连续的) range of cows from the milking lineup(阵容) to play the game. However, for all the cowsto have fun they should not differ too much in height.

Farmer John has madea list of Q (1 ≤ Q ≤ 200,000) potential(潜在的) groups of cows and their heights (1 ≤ height ≤ 1,000,000). For each group, he wants your help todetermine the difference in height between the shortest and the tallest cow inthe group.

Input

Line 1:Two space-separated integers(整数), N and Q. 
Lines 2..N+1: Line i+1contains a single integer(整数) 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(整数) that is a response to a reply and indicates(表明) the difference in height between the tallestand shortest cow in the range.

SampleInput

6 3

1

7

3

4

2

5

1 5

4 6

2 2

SampleOutput

6

3

0

Source

USACO2007 January Silver

线段树解法

Problem: 3264

User: ksq2013

Memory: 1692K

Time: 3891MS

Language: G++

Result: Accepted

 

#include<cstdio>
#include<cstdlib>
#include<iostream>
using namespace std;
int n,q,_max[200001],_min[200001];
void update(int s,int t,int k,int add,int p)
{
	if(!(s^t)){
		_max[k]=_min[k]=p;
		return;
	}int m=(s+t)>>1;
	if(add<=m)update(s,m,k<<1,add,p);
	else update(m+1,t,k<<1|1,add,p);
	_max[k]=max(_max[k<<1],_max[k<<1|1]);
	_min[k]=min(_min[k<<1],_min[k<<1|1]);
}
int query1(int s,int t,int k,int l,int r)
{
	if(l<=s&&t<=r)return _max[k];
	int m=(s+t)>>1,res1=0,res2=0;
	if(l<=m)res1=query1(s,m,k<<1,l,r);
	if(m<r)res2=query1(m+1,t,k<<1|1,l,r);
	return max(res1,res2);
}
int query2(int s,int t,int k,int l,int r)
{
	if(l<=s&&t<=r)return _min[k];
	int m=(s+t)>>1,res1=0x7fffffff,res2=0x7fffffff;
	if(l<=m)res1=query2(s,m,k<<1,l,r);
	if(m<r)res2=query2(m+1,t,k<<1|1,l,r);
	return min(res1,res2);
}
int main()
{
	scanf("%d%d",&n,&q);
	for(int i=1;i<=n;i++){
		int x;
		scanf("%d",&x);
		update(1,n,1,i,x);
	}
	for(int i=1;i<=q;i++){
		int a,b;
		scanf("%d%d",&a,&b);
		printf("%d\n",query1(1,n,1,a,b)-query2(1,n,1,a,b));
	}
	return 0;
}