HDU - 5443The Water Problem 线段树（区间查询最大值）

Posted 2023-02-07

The Water Problem

Time Limit: 1500/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 3386    Accepted Submission(s): 2416

 

Problem Description

In Land waterless, water is a very limited resource. People always fight for the biggest source of water. Given a sequence of water sources with a1,a2,a3,...,anrepresenting the size of the water source. Given a set of queries each containing 2 integers l and r, please find out the biggest water source between al and ar.

 

 

Input

First you are given an integer T(T≤10) indicating the number of test cases. For each test case, there is a number n(0≤n≤1000) on a line representing the number of water sources. n integers follow, respectively a1,a2,a3,...,an, and each integer is in 1,...,106. On the next line, there is a number q(0≤q≤1000) representing the number of queries. After that, there will be q lines with two integers l and r(1≤l≤r≤n) indicating the range of which you should find out the biggest water source.

 

 

Output

For each query, output an integer representing the size of the biggest water source.

 

 

Sample Input

 

3 1 100 1 1 1 5 1 2 3 4 5 5 1 2 1 3 2 4 3 4 3 5 3 1 999999 1 4 1 1 1 2 2 3 3 3

 

 

Sample Output

 

100 2 3 4 4 5 1 999999 999999 1

 

 

Source

​​2015 ACM/ICPC Asia Regional Changchun Online​​

 

 

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#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <cstring>
#define N 200010
#define ll long long
using namespace std; 
ll a[N];
int n,m; 
struct node  //线段树

    int l,r;
    ll dat;
tree[N<<2]; 
void PushUp(int x)   //线段树维护结点信息

    tree[x].dat=max(tree[x<<1].dat,tree[x<<1|1].dat);

void build(int x,int l,int r)  //线段树建树

    tree[x].l=l;
    tree[x].r=r;
    if (l==r)   //叶子节点
    
        tree[x].dat=a[l];
        return;
    
    int mid=(l+r)>>1;
    build(x<<1,l,mid);
    build(x<<1|1,mid+1,r);
    PushUp(x);
  
ll query(int x,int l,int r)  //线段树区间查询
  
    if (l<=tree[x].l&&r>=tree[x].r) 
        return abs(tree[x].dat);  //找到
        
    int mid=(tree[x].l+tree[x].r)>>1;
    
    ll ans=0;
    if (l<=mid) ans=max(ans,query(x<<1,l,r));
    if (r>mid)  ans=max(ans,query(x<<1|1,l,r)); 
    return ans;
 
void update(int x,int y,ll k)  //线段树单点修改

    if (tree[x].l==tree[x].r) 
    
        tree[x].dat = k;
        return ;
    
    int mid=(tree[x].l+tree[x].r)>>1;
    if (y<=mid)  update(x<<1,y,k);
    else         update(x<<1|1,y,k);
    PushUp(x);

 
int main()

    int t;
    cin>>t;
     while(t--)
     
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
            scanf("%lld",&a[i]);
            scanf("%d",&m);
        build(1,1,n);
        char s[10];
        while(m--)
        
            
            int i,j;
         
                scanf("%d%d",&i,&j);
                printf("%lld\\n",query(1,i,j));
                
        
     
    return 0;