POJ 3264 线段树

Balanced Lineup

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 16163		Accepted: 7468
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

题意很简单，就不翻译了。

此题数据量非常大，简单的模拟必然会超时，线段树是个不错的选择

我们可以通过线段树分别查找区间最大值和区间最小值，之差就是答案了。。

Source Code

Problem: 3264		User: bingshen
Memory: 2384K		Time: 1954MS
Language: C++		Result: Accepted

• Source Code

  #include<stdio.h>
  #include<string.h>
  
  struct node
  {
  	int max;
  	int min;
  	int l;
  	int r;
  };
  node tree[200000];
  int h[50005];
  int max,min;
  
  int MAX(int a,int b)
  {
  	if(a>b)
  		return a;
  	else
  		return b;
  }
  
  int MIN(int a,int b)
  {
  	if(a>b)
  		return b;
  	else
  		return a;
  }
  
  void build(int l,int r,int root)
  {
  	tree[root].l=l;
  	tree[root].r=r;
  	if(l==r)
  	{
  		tree[root].max=h[l];
  		tree[root].min=h[l];
  		return;
  	}
  	int mid=(l+r)>>1;
  	build(l,mid,root*2);
  	build(mid+1,r,root*2+1);	
  	tree[root].max=MAX(tree[2*root].max,tree[2*root+1].max);
  	tree[root].min=MIN(tree[2*root].min,tree[2*root+1].min);
  }
  
  void findmax(int l,int r,int root)
  {
  	if(tree[root].l==l&&tree[root].r==r)
  	{
  		if(tree[root].max>max)
  			max=tree[root].max;
  		return;
  	}
  	int mid=(tree[root].l+tree[root].r)>>1;
  	if(mid>=r)
  		findmax(l,r,root*2);
  	else if(mid<l)
  		findmax(l,r,root*2+1);
  	else 
  	{
  		findmax(l,mid,root*2);
  		findmax(mid+1,r,root*2+1);
  	}
  }
  
  void findmin(int l,int r,int root)
  {
  	if(tree[root].l==l&&tree[root].r==r)
  	{
  		if(tree[root].min<min)
  			min=tree[root].min;
  		return;
  	}
  	int mid=(tree[root].l+tree[root].r)>>1;
  	if(mid>=r)
  		findmin(l,r,root*2);
  	else if(mid<l)
  		findmin(l,r,root*2+1);
  	else 
  	{
  		findmin(l,mid,root*2);
  		findmin(mid+1,r,root*2+1);
  	}
  }
  
  int main()
  {
  	int n,q,i,a,b;
  	scanf("%d%d",&n,&q);
  	for(i=1;i<=n;i++)
  		scanf("%d",&h[i]);
  	build(1,n,1);
  	while(q--)
  	{
  		max=0;
  		min=99999999;
  		scanf("%d%d",&a,&b);
  		findmax(a,b,1);
  		findmin(a,b,1);
  		printf("%d/n",max-min);
  	}
  	return 0;
  }

• 果断的1Y，很久没这个感觉了。。