E - Balanced Lineup POJ - 3264

本文介绍了一种使用线段树数据结构来高效解决区间内最大高度差问题的方法。具体涉及如何构建线段树、更新节点信息以及查询指定区间内的最大与最小值,最终求得最大高度差。

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

简单的线段树:

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn=201000;
const int inf=0x3f3f3f3f;
#define lson rt<<1,l,mid
#define rson rt<<1|1,mid+1,r
struct node
{
	int minn,maxx,l,r,val;
}tree[maxn*4];
void pushup(int rt)
{
	tree[rt].minn=min(tree[rt<<1].minn,tree[rt<<1|1].minn);
	tree[rt].maxx=max(tree[rt<<1].maxx,tree[rt<<1|1].maxx);
}
void build(int rt,int l,int r)
{
	tree[rt].l=l;
	tree[rt].r=r;
	if(l==r)
	{
		scanf("%d",&tree[rt].val);
		tree[rt].minn=tree[rt].val;
		tree[rt].maxx=tree[rt].val;
		return;
	}
	int mid=(l+r)/2;
	build(lson);
	build(rson);
	pushup(rt);
}
int high=0,low=inf;
void query(int rt,int l,int r)
{
	if(tree[rt].l==l&&tree[rt].r==r)
	{
		high=max(high,tree[rt].maxx);
		low=min(low,tree[rt].minn);
		return;
	}
	int mid=(tree[rt].l+tree[rt].r)/2;
	if(l>mid)
	{
		query(rt<<1|1,l,r);
	}
	else if(mid>=r)
	{
		query(rt<<1,l,r);
	}
	else
	{
		query(lson);
		query(rson);
	}
}
int main ()
{
	int n,m;
	scanf("%d%d",&n,&m);
	build(1,1,n);
	while(m--)
	{
		int xx,yy;
		high=0;
		low=inf;
		scanf("%d%d",&xx,&yy);
		query(1,xx,yy);
		printf("%d\n",high-low);
	}
} 
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