Balanced Lineup（点击转到）
| Time Limit: 5000MS | Memory Limit: 65536K | |
| Total Submissions: 64368 | Accepted: 30003 | |
| 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
1.题目含义：
给定Q (1 ≤ Q ≤ 200,000)个数A1,A2 … AQ,，多次求任一区间Ai – Aj中最大数和最小数的差。
2.线段树：
2.1线段树适用于和区间统计有关的问题，当题目中有 “区间”，‘’任意“，多次”等词汇的时候，便要考虑用线段树分区间进行求解。
2.2线段树的过程就分为：建线段树——插入数据——查询更新的一个过程。
3.注意：
用指针的形式还是用数组形式（下标从零开始）
#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
#define inf 0x3f3f3f3f
int nmax=-inf;
int nmin=inf;
struct node
{
int l,r;
int nmax;
int nmin;
int mid()
{
return (l+r)/2;
}
}tree[800010];
void buildTree(int root ,int l,int r)//root当前节点编号在数组中的下标（从0开始）
{
tree[root].l=l;
tree[root].r=r;
tree[root].nmax=-inf;
tree[root].nmin=inf;
int mid=(l+r)/2;
if(l!=r)
{
buildTree(2*root+1,l,mid);
buildTree(2*root+2,mid+1,r);
}
}
void insert(int root,int i,int v)//将第i个数，其值为v，插入到线段树编号为root的位置
{
if(tree[root].l==tree[root].r)
{
tree[root].nmax=tree[root].nmin=v;
return ;
}
tree[root].nmax=max(tree[root].nmax,v);
tree[root].nmin=min(tree[root].nmin,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)
{
if(tree[root].nmin>=nmin&&tree[root].nmax<=nmax)//最优性剪枝
return ;
if(tree[root].l==s&&tree[root].r==e)
{
nmin=min(tree[root].nmin,nmin);
nmax=max(tree[root].nmax,nmax);
return ;
}
if(e<=tree[root].mid())//在左边
{
query(2*root+1,s,e);
}
else if(s>tree[root].mid())//在右边
{
query(2*root+2,s,e);
}
else//在中间
{
query(2*root+1,s,tree[root].mid());
query(2*root+2,tree[root].mid()+1,e);
}
}
int main()
{
int m,n;
scanf("%d%d",&n,&m);
buildTree(0,1,n);
for(int i=1;i<=n;i++)
{
int num;
scanf("%d",&num);
insert(0,i,num);
}
for(int i=0;i<m;i++)
{
int s;
int e;
scanf("%d%d",&s,&e);
nmin=inf;
nmax=-inf;
query(0,s,e);
printf("%d\n",nmax-nmin);
}
return 0;
}