[线段树]USACO07JAN 平衡的阵容Balanced Lineup

题目传送门：https://www.luogu.org/problemnew/show/P2880

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题目描述
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.

每天,农夫 John 的N(1 <= N <= 50,000)头牛总是按同一序列排队. 有一天, John 决定让一些牛们玩一场飞盘比赛. 他准备找一群在对列中为置连续的牛来进行比赛. 但是为了避免水平悬殊,牛的身高不应该相差太大. John 准备了Q (1 <= Q <= 180,000) 个可能的牛的选择和所有牛的身高 (1 <= 身高 <= 1,000,000). 他想知道每一组里面最高和最低的牛的身高差别.

输入格式：
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.

第1行：N,Q;
第2到N+1行：每头牛的身高;
第N+2到N+Q+1行：两个整数A和B，表示从A到B的所有牛。（1<=A<=B<=N）

输出格式：
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.

每行输出一个数，为最大数与最小数的差。

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题解：

“区间求某个值” 这一类的问题首选解法当然是线段树和树状数组。
题目要让我们求一个区间内两个元素最大差，看似无从下手，但是分析一下可以得出最大差不就是最大值减去最小值吗？【惊讶】
那么，这道题就变成一道线段树模板题了。
维护一个线段树，要求查询每个区间的最大值和最小值。

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代码：

#include
#include
#include
#define maxn 60000
#define inf 1000010
using namespace std;
struct node
{
    int l,r,s,t;
    // s=shortest,t=tallest
};
node tree[maxn*4+10];
int a[maxn],n,q;
void build(int p,int x,int y)
{
    //printf("%d %d\n",x,y); 
    int mid=(x+y)>>1;
    tree[p].l=x;
    tree[p].r=y;
    if(x1,x,mid);
        build(p<<1|1,mid+1,y);
        tree[p].s=min(tree[p<<1].s,tree[p<<1|1].s);
        tree[p].t=max(tree[p<<1].t,tree[p<<1|1].t);
    }
    else
    {
        tree[p].s=tree[p].t=a[x];
        //printf("%d ",a[x]);
    }
}
int queryMax(int p,int x,int y)
{
    if(x<=tree[p].l&&tree[p].r<=y)
    {
        return tree[p].t;
    }
    int mid=(tree[p].l+tree[p].r)>>1,lmax=0,rmax=0;
    if(x<=mid&&y>=tree[p].l)
    {
        lmax=queryMax(p<<1,x,y);
    }
    if(y>mid&&x<=tree[p].r)
    {
        rmax=queryMax(p<<1|1,x,y);
    }
    return max(lmax,rmax);
}

int queryMin(int p,int x,int y)
{
    if(x<=tree[p].l&&tree[p].r<=y)
    {
        return tree[p].s;
    }
    int mid=(tree[p].l+tree[p].r)>>1,lmin=inf,rmin=inf;
    if(x<=mid&&y>=tree[p].l)
    {
        lmin=queryMin(p<<1,x,y);
    }
    if(y>mid&&x<=tree[p].r)
    {
        rmin=queryMin(p<<1|1,x,y);
    }
    return min(lmin,rmin);
}
int main()
{
    scanf("%d%d",&n,&q);
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&a[i]);
    }
    build(1,1,n);
    while(q--)
    {
        int x,y;
        scanf("%d%d",&x,&y);
        printf("%d\n",queryMax(1,x,y)-queryMin(1,x,y));
    }
    return 0;
}