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

 1 #include<cstdio>
 2 #include<algorithm>
 3 #include<set>
 4 #include<vector>
 5 #include<cstring>
 6 #include<iostream>
 7 using namespace std;
 8 const int maxn=5e4+5;
 9 int a[maxn];
10 struct Node{
11     int l,r,maxnum,minnum;
12 }tree[maxn*4];
13 void push_up(int x){
14     tree[x].maxnum=max(tree[x<<1].maxnum,tree[x<<1|1].maxnum);
15     tree[x].minnum=min(tree[x<<1].minnum,tree[x<<1|1].minnum);
16 }
17 int  query1(int x,int l,int r){
18     int L=tree[x].l,R=tree[x].r;
19     if(l<=L&&R<=r){
20         return tree[x].maxnum;
21     }
22     else {
23         int mid=L+R>>1;
24         int ans=0;
25         if(l<=mid)ans=max(ans,query1(x<<1,l,r));
26         if(mid<r)ans=max(ans,query1(x<<1|1,l,r));
27         return ans;
28     }
29 }
30 int  query2(int x,int l,int r){
31     int L=tree[x].l,R=tree[x].r;
32     if(l<=L&&R<=r){
33         return tree[x].minnum;
34     }
35     else {
36         int mid=L+R>>1;
37         int ans=1e7;
38         if(l<=mid)ans=min(ans,query2(x<<1,l,r));
39         if(mid<r)ans=min(ans,query2(x<<1|1,l,r));
40         return ans;
41     }
42 }
43 void build(int x,int l,int r){
44     tree[x].l=l,tree[x].r=r;
45     if(l==r){
46         tree[x].maxnum=tree[x].minnum=a[l];
47     }
48     else {
49         int mid=l+r>>1;
50         build(x<<1,l,mid);
51         build(x<<1|1,mid+1,r);
52         push_up(x);
53     }
54 }
55 int main(){
56     int n,q;
57     scanf("%d%d",&n,&q);
58     for(int i=1;i<=n;i++)scanf("%d",&a[i]);
59     build(1,1,n);
60     for(int i=1;i<=q;i++){
61         int x,y;
62         scanf("%d%d",&x,&y);
63         printf("%d\n",query1(1,x,y)-query2(1,x,y));
64     }
65     return 0;
66 }