poj 3246 Balanced Lineup（线段树）

Balanced Lineup

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 38942		Accepted: 18247
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 andQ.

Lines 2..N+1: Line i+1 contains a single integer that is the height of cowi

Lines N+2..N+Q+1: Two integers A and B (1 ≤A ≤
B ≤ N), representing the range of cows from A toB 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



题目大意：在一定区间内给定一些数。要求求出在某一区间内最大值和最小值的差。

线段树的题目。对于这道题目，既然是求最大值和最小值的差，那么必定要在区间里面存放最大值和最小值。同一时候这道题目仅仅是单纯的要求查询区间内的差值，不须要进行更新。

#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下面的节点所有更新。

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;
}