Balanced Lineup(線段樹-樹狀數組)

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Balanced Lineup

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 55919		Accepted: 26205
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

6 3
1
7
3
4
2
5
1 5
4 6
2 2

Sample Output

6
3
0

Source

USACO 2007 January Silver

Select Code

#include<iostream>
using namespace std;
const int INF = 0xffffff0;
int minV = INF;
int maxV = -INF;
struct Node		//不要左右子節點指針的做法 
{
	int L,R;
	int minV,maxV;
	int Mid()
	{
		return (L+R)/2;
	}
};
Node tree[800010];	//4倍葉子節點數量就夠 

void BuildTree(int root, int L, int R)
{
	tree[root].L = L;
	tree[root].R = R;
	tree[root].minV = INF;
	tree[root].maxV = -INF;
	if(L!=R)
	{
		BuildTree(2*root+1, L, (L+R)/2);
		BuildTree(2*root+2, (L+R)/2+1, R);
	}
}

void Insert(int root, int i, int v)
//將第i個數，其值爲v，插入線段樹 
{
	if(tree[root].L == tree[root].R)	//成立則亦有tree[root].R==i 
	{
		tree[root].minV = tree[root].maxV = v;
		return ;
	}
	tree[root].minV = min(tree[root].minV , v);//終於找到bug了 
	tree[root].maxV = max(tree[root].maxV , v);
	if(i <= tree[root].Mid())
		Insert(2*root + 1, i, v);
	else
		Insert(2*root + 2, i, v);		
}

void Query(int root, int s, int e)
//查詢區間[s,e]中的最大值和最小值，如果更優就記錄在全局變量裏 
{
	if( tree[root].minV >= minV && tree[root].maxV <= maxV )
		return;
	if( tree[root].L == s && tree[root].R == e )
	{
			minV = min( minV, tree[root].minV );
//			minV = min( minV, tree[root].minV );
			maxV = max( maxV, tree[root].maxV );
			return ;
	}
	if( e <= tree[root].Mid() )
		Query( 2*root + 1, s, e );
	else if( s > tree[root].Mid() )
		Query( 2*root + 2, s, e );
	else
	{
		Query( 2*root + 1, s, tree[root].Mid() );
		Query( 2*root + 2, tree[root].Mid() + 1, e );
		
	}
	
}

int main()
{
	int n,q,h;
	int i,j,k;
	scanf("%d%d",&n,&q);
	BuildTree(0, 1, n);
	for(i=1; i<=n; i++)
	{
		scanf("%d",&h);
		Insert(0, i, h);
	}
	for(i=0; i<q; i++)
	{
		int s,e;
		scanf("%d%d",&s,&e);
		minV=INF;
		maxV=-INF;
		Query(0, s, e);
		printf("%d\n",maxV-minV);
	}
	return 0;
}