POJ 3264 Balanced Lineup 線段樹

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

6 3
1
7
3
4
2
5
1 5
4 6
2 2Sample Output

6
3
0

#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <climits>
using namespace std;
/*
int max(int a,int b){
	return a>b?a:b;
}*/
const int MAXN=50010; 
struct tnode {
	int l,r,mmax,mmin;
};
tnode node[MAXN*4];
int a[MAXN],maxv,minv;
void init(int k,int l,int r) // 節點k對應的區間為 [l,r) 
{
	node[k].l=l;
	node[k].r=r;
	if(r-l==1){          // 區間只有一個元素的情況 
		node[k].mmax=node[k].mmin=a[l];
		return ;
	} 
	int mid=(l+r)/2;
	init(2*k+1,l,mid);
	init(2*k+2,mid,r); //分別構造左右子樹 
	node[k].mmax=max(node[2*k+1].mmax,node[2*k+2].mmax);
	node[k].mmin=min(node[2*k+1].mmin,node[2*k+2].mmin);
}
void query(int l,int r,int k)
{
	if(node[k].l==l && node[k].r==r ){ // 查詢的區間和節點區間完全重合 
		maxv=max(maxv,node[k].mmax);
		minv=min(minv,node[k].mmin);
		return ;
	}
	int mid=(node[k].l+node[k].r)/2;
	if(mid<=l) query(l,r,2*k+2);     //所查詢的區間在右子樹 
	else if(mid>=r) query(l,r,2*k+1); //所查詢的區間在左子樹 
	else {
		query(l,mid,2*k+1);   //所查詢的區間分別在左右子樹上 
		query(mid,r,2*k+2);
	}
}
int main()
{
	int n,q;
	while(cin>>n>>q)
	{
		memset(node,0,sizeof(node));
		for(int i=0;i<n;i++) 
			scanf("%d",&a[i]);
		init(0,0,n);   //初始化線段樹 
		for(int i=0;i<q;i++)
		{
			int x,y;
			scanf("%d %d",&x,&y);
			maxv=0;minv=INT_MAX;
			query(x-1,y,0);    //這裡是左閉右開區間 
			cout<<maxv-minv<<endl;
		}
	}
	return 0;
}