Number sequence interval maximum value | line segment tree

Maximum value of sequence interval

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acwing 1270

Time limit：2s

Memory limit：64MB

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Problem Description

Enter a string of numbers and give you M queries. Each query will give you two numbers X, Y asks you to say the maximum number in the interval from X to Y.

Input

• The two integers N and M in the first line represent the number of digits and the number of queries;

  The next row is N numbers;

  The next M lines, each line has two integers X, Y.

  1≤N≤10 ⁵
  1≤M≤10 ⁶
  1≤X≤Y≤N No
  number in the sequence exceeds 2 ³¹ −1

Output

There are a total of M lines of output, and each line outputs a number.

Sample Input

10 2
3 2 4 5 6 8 1 2 9 7
1 4
3 8

Sample Output

5
8

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Line segment tree solution

AC code:

#include<cstdio>
#include<iostream>
using namespace std;
int n,m,x[100005];          //原数组
struct Node{
    
    
    int l,r,w;              //线段树的左、右、值
};
Node tree[400005];          //线段树
void build(int num,int l,int r){
    
    
    if(l == r){
    
                 //相等说明已经到了叶子节点
        tree[num].l = l,tree[num].r = r,tree[num].w = x[l];return;
    }
    int mid = (l + r) >> 1;
    int lc = num << 1;      //乘2是左子树
    int rc = num << 1 | 1;  //乘2或1是右子树
    build(lc,l,mid);        //建左子树
    build(rc,mid + 1,r);  //建右子树
    tree[num].l = l,tree[num].r = r,tree[num].w = max(tree[lc].w,tree[rc].w);return;    //保存左右子树的最大子（区间最大）
}
int find(int num,int l,int r){
    
    
    if(tree[num].l == l && tree[num].r == r)
        return tree[num].w;             //找到目标区间
    int mid = (tree[num].l + tree[num].r) >> 1;
    int lc = num << 1;
    int rc = num << 1 | 1;
    if(l > mid)
        return find(rc,l,r);    //继续在右子树找
    else if(r <= mid)
        return find(lc,l,r);    //继续在左子树找
    else return max(find(lc,l,mid),find(rc,mid + 1,r)); //左子树找一段，右子树找一段
}
int main(){
    
    
    scanf("%d %d",&n,&m);
    for(int i = 1;i <= n;++i)
        scanf("%d",x + i);
    build(1,1,n);
    while(m--){
    
    
        int l,r;
        scanf("%d %d",&l,&r);
        printf("%d\n",find(1,l,r));
    }
    return 0;
}