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.

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

题解：线段树的模板题，挺简单的;

#include <iostream>
#include <cstdio>
using namespace std;
#define inf 0x3f3f3f3f
const int maxn=50005;
int n, q, x, y, ansmax, ansmin;
struct seg
{
    int l,r,ma,mi;
}tree[maxn<<2];
void build(int p, int l, int r)
{
    tree[p].l=l;
    tree[p].r=r;
    tree[p].ma=-inf;
    tree[p].mi=inf;
    if(l==r) return;
    int mid=(l+r)>>1;
    build(p<<1,l,mid);
    build(p<<1|1,mid+1,r);
}
void add(int p, int x, int y)
{
    if(tree[p].l==tree[p].r)
    {
        tree[p].ma=y;
        tree[p].mi=y;
        return;
    }
    int mid=(tree[p].l+tree[p].r)>>1;
    if(x<=mid) add(p<<1,x,y);
    else add(p<<1|1,x,y);
    tree[p].ma=max( tree[p<<1].ma, tree[p<<1|1].ma );
    tree[p].mi=min( tree[p<<1].mi, tree[p<<1|1].mi );
}
int askmax(int p, int l, int r)
{
    if(l==tree[p].l && r==tree[p].r) return max(ansmax, tree[p].ma );
    int mid=(tree[p].l+tree[p].r)>>1;
    if(r<=mid) return max(ansmax, askmax(p<<1, l, r) );
    else if(l>mid) return max(ansmax, askmax(p<<1|1, l, r) );
    else return max(ansmax, max(askmax(p<<1,l,mid), askmax(p<<1|1,mid+1,r) ) );
}//找区间内最大值;
int askmin(int p, int l, int r)
{
    if(l==tree[p].l && r==tree[p].r) return min(ansmin, tree[p].mi );
    int mid=(tree[p].l+tree[p].r)>>1;
    if(r<=mid) return min(ansmin, askmin(p<<1, l, r) );
    else if(l>mid) return min(ansmin, askmin(p<<1|1, l, r) );
    else return min(ansmin, min(askmin(p<<1,l,mid), askmin(p<<1|1,mid+1,r) ) );
}//找区间内最小值;
int main()
{
    while(cin>>n>>q)
    {
        build(1,1,n);
        for(int i=1; i<=n; i++)
            scanf("%d",&x), add(1,i,x);
        for(int i=1; i<=q; i++)
        {
            scanf("%d%d",&x,&y);
            ansmax=-inf;
            ansmin=inf;
            printf("%d\n",askmax(1,x,y)-askmin(1,x,y));
        }
    }
    return 0;
}