POJ3264 Balances Lineup

建两颗线段树分别存最大和最小值，模板题~

#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
const int maxn=1e6+;
struct node {
    int l;
    int r;
    int sum;
}segTree1[maxn*],segTree2[maxn*];
int a[maxn];
int N;
int Q;
int x,y;
void build1 (int i,int l,int r) {
    segTree1[i].l=l;
    segTree1[i].r=r;
    if (l==r) {
        segTree1[i].sum=a[l];
        return;
    }
    int mid=(l+r)>>;
    build1(i<<,l,mid);
    build1(i<<|,mid+,r);
    segTree1[i].sum=max(segTree1[i<<].sum,segTree1[i<<|].sum);
}
int query1 (int i,int l,int r) {
    if (l==segTree1[i].l&&r==segTree1[i].r) return segTree1[i].sum;
    int mid=(segTree1[i].l+segTree1[i].r)>>;
    if (r<=mid) return query1(i<<,l,r);
    else if (l>mid) return query1(i<<|,l,r);
    else return max(query1(i<<,l,mid),query1(i<<|,mid+,r));
}
void build2 (int i,int l,int r) {
    segTree2[i].l=l;
    segTree2[i].r=r;
    if (l==r) {
        segTree2[i].sum=a[l];
        return;
    }
    int mid=(l+r)>>;
    build2(i<<,l,mid);
    build2(i<<|,mid+,r);
    segTree2[i].sum=min(segTree2[i<<].sum,segTree2[i<<|].sum);
}
int query2 (int i,int l,int r) {
    if (l==segTree2[i].l&&r==segTree2[i].r) return segTree2[i].sum;
    int mid=(segTree2[i].l+segTree2[i].r)>>;
    if (r<=mid) return query2(i<<,l,r);
    else if (l>mid) return query2(i<<|,l,r);
    else return min(query2(i<<,l,mid),query2(i<<|,mid+,r));
}
int main () {
    scanf ("%d %d",&N,&Q);
    for (int i=;i<=N;i++) scanf ("%d",&a[i]);
    build1(,,N);
    build2(,,N);
    for (int i=;i<Q;i++) {
        scanf ("%d %d",&x,&y);
        printf ("%d\n",query1(,x,y)-query2(,x,y));
    }
    return ;
}