Balanced Lineup——区间最大值区间最小值

Posted 2022-12-17 j666

题目链接

题意:

给定n个数，q次询问，查询当前区间最大值与最小值的差值

题解：

线段树模板题

区间查询最大值、最小值

代码：

#include<iostream>
#include<stdio.h>
#include<math.h>
#include<algorithm>
#include<vector>
#include<map>
using namespace std;
typedef long long ll;
const int maxn=1e5+5;
struct node

    int l,r;
    int minn,maxx;
 tree[maxn<<2];
int n,a[maxn],q;
void push_up(int x)

    tree[x].maxx=max(tree[x<<1].maxx,tree[x<<1|1].maxx);
    tree[x].minn=min(tree[x<<1].minn,tree[x<<1|1].minn);

void build(int x,int l,int r)

    tree[x].l=l,tree[x].r=r;
    if(l==r)
    
        tree[x].maxx=tree[x].minn=a[l];
    
    else
    
        int mid=(l+r)>>1;
        build(x<<1,l,mid);
        build(x<<1|1,mid+1,r);
        push_up(x);
    

ll query(int x,int l,int r,int flag)

    int L=tree[x].l,R=tree[x].r;

    if(l<=L && R<=r)
    
        if(flag)return tree[x].maxx;
        else return tree[x].minn;
    
    else
    

        if(flag)
        
            ll ans=0;
            int mid=(L+R)>>1;
            if(mid>=l)ans=max(ans,query(x<<1,l,r,flag));
            if(mid<r)ans=max(ans,query(x<<1|1,l,r,flag));
            push_up(x);
            return ans;
        
        else
        
            ll ans=0x3f3f3f3f;
            int mid=(L+R)>>1;
            if(mid>=l)ans=min(ans,query(x<<1,l,r,flag));
            if(mid<r)ans=min(ans,query(x<<1|1,l,r,flag));
            push_up(x);
            return ans;
        

    


int main()

    scanf("%d%d",&n,&q);
    for(int i=1; i<=n; i++)
        scanf("%d",&a[i]);
    build(1,1,n);
    for(int i=1; i<=q; i++)
    
        int l,r;
        scanf("%d%d",&l,&r);
        ll ans=query(1,l,r,1)-query(1,l,r,0);
        printf("%lld\n",ans);
    
    return 0;

View Code