G - Balanced Lineup POJ - 3264 线段树最大最小值区间查询模版题

题意 给出一个序列  每次查询区间的max-min是多少

思路：直接维护max 和min即可  写两个query分别查最大最小值

 #include<cstdio>
 #include<algorithm>
 #include<set>
 #include<vector>
 #include<cstring>
 #include<iostream>
 using namespace std;
 const int maxn=5e4+;
 int a[maxn];
 struct Node{
     int l,r,maxnum,minnum;
 }tree[maxn*];
 void push_up(int x){
     tree[x].maxnum=max(tree[x<<].maxnum,tree[x<<|].maxnum);
     tree[x].minnum=min(tree[x<<].minnum,tree[x<<|].minnum);
 }
 int  query1(int x,int l,int r){
     int L=tree[x].l,R=tree[x].r;
     if(l<=L&&R<=r){
         return tree[x].maxnum;
     }
     else {
         int mid=L+R>>;
         int ans=;
         if(l<=mid)ans=max(ans,query1(x<<,l,r));
         if(mid<r)ans=max(ans,query1(x<<|,l,r));
         return ans;
     }
 }
 int  query2(int x,int l,int r){
     int L=tree[x].l,R=tree[x].r;
     if(l<=L&&R<=r){
         return tree[x].minnum;
     }
     else {
         int mid=L+R>>;
         int ans=1e7;
         if(l<=mid)ans=min(ans,query2(x<<,l,r));
         if(mid<r)ans=min(ans,query2(x<<|,l,r));
         return ans;
     }
 }
 void build(int x,int l,int r){
     tree[x].l=l,tree[x].r=r;
     if(l==r){
         tree[x].maxnum=tree[x].minnum=a[l];
     }
     else {
         int mid=l+r>>;
         build(x<<,l,mid);
         build(x<<|,mid+,r);
         push_up(x);
     }
 }
 int main(){
     int n,q;
     scanf("%d%d",&n,&q);
     for(int i=;i<=n;i++)scanf("%d",&a[i]);
     build(,,n);
     for(int i=;i<=q;i++){
         int x,y;
         scanf("%d%d",&x,&y);
         printf("%d\n",query1(,x,y)-query2(,x,y));
     }
     return ;
 }