poj 3246 Balanced Lineup（线段树）

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.

#include<stdio.h>
#include<string.h>
#define max(a,b) a>b?a:b
#define min(a,b) a<b?a:b
#define INF 99999999
#define N 50005
struct tree{
	int l,r,maxi,mini;
	int mid(){
		return l+r>>1;
	}
}tree[N<<2];
int ma=-INF,mi=INF;
void build(int l,int r,int root)
{
	tree[root].l=l;
	tree[root].r=r;
	tree[root].maxi=-INF;
	tree[root].mini=INF;  //初始化最大最小值
	if(l==r){
	
		return;
	}
	int mid=l+r>>1;
	build(l,mid,root<<1);
	build(mid+1,r,root<<1|1);
}
void update(int i,int z,int root)
{

	if(tree[root].l==tree[root].r){
		tree[root].mini=tree[root].maxi=z;
		return;
	}
	tree[root].maxi=max(tree[root].maxi,z);
	tree[root].mini=min(tree[root].mini,z);      //每次都更新最大和最小值
    if(i<=tree[root].mid())update(i,z,root<<1);   //这里将i以下的节点全部更新。而i与mid 是有关系的。 
    else update(i,z,root<<1|1);
}
void Query(int l,int r,int root)
{
	if(tree[root].mini>=mi&&tree[root].maxi<=ma)return;
   if(l==tree[root].l&&r==tree[root].r){
   	mi=min(mi,tree[root].mini);
   	ma=max(ma,tree[root].maxi);
   	return;
   }
   int mid=tree[root].l+tree[root].r>>1;
   if(r<=mid){
   	Query(l,r,root<<1);
   }
   else if(l>mid){
   	Query(l,r,root<<1|1);
   }
   else {
   	Query(l,mid,root<<1);
   	Query(mid+1,r,root<<1|1);
   }
   return ;
}
int main()
{
	int n,Q,cow[200005],a,b;
	int i,j,k;
	while(scanf("%d%d",&n,&Q)!=EOF)
	{
		build(1,n,1);
		for(i=1;i<=n;i++)
		{
		scanf("%d",&cow[i]);
	   update(i,cow[i],1);     //对于第i个数字进行插入
	}

	while(Q--)
	{
		scanf("%d%d",&a,&b);
		ma=-INF;
		mi=INF;
		Query(a,b,1);
		printf("%d\n",ma-mi);
	}
	}
	return 0;
}