2017 ACM-ICPC 北京网络赛 Minimum 线段树

描述

You are given a list of integers a0, a1, …, a2^k-1.

You need to support two types of queries:

1. Output Minx,y∈[l,r] {ax∙ay}.

2. Let ax=y.

输入

The first line is an integer T, indicating the number of test cases. (1≤T≤10).

For each test case:

The first line contains an integer k (0 ≤ k ≤ 17).

The following line contains 2k integers, a0, a1, …, a2^k-1 (-2k ≤ ai < 2k).

The next line contains a integer  (1 ≤ Q < 2k), indicating the number of queries. Then next Q lines, each line is one of:

1. 1 l r: Output Minx,y∈[l,r]{ax∙ay}. (0 ≤ l ≤ r < 2k)

2. 2 x y: Let ax=y. (0 ≤ x < 2k, -2k ≤ y < 2k)

输出

For each query 1, output a line contains an integer, indicating the answer.

样例输入

1
3
1 1 2 2 1 1 2 2
5
1 0 7
1 1 2
2 1 2
2 2 2
1 1 2

样例输出

1
1
4

一道线段树单点更新模版题，比赛的时候居然因为N定义小了WA，可是我明明放大了呀，摊手jpg。

#include<iostream>
#include<math.h>
#include<algorithm>
#define ll long long
#define MAX_N 132000

using namespace std;

struct node
{
    ll minn,maxx;
    int l,r;
}num[MAX_N*4];
int pp[MAX_N];
void pushUp(int count)
{
    num[count].minn = min(num[count<<1].minn,num[count<<1|1].minn);
    num[count].maxx = max(num[count<<1].maxx,num[count<<1|1].maxx);
}
void built_tree(int l,int r,int count)
{
    num[count].l=l;
    num[count].r=r;
    if(l==r)
    {
        num[count].minn = pp[l];
        num[count].maxx = pp[l];
        return ;
    }
    int mid = (l+r)/2;
    built_tree(l, mid, count*2);
    built_tree(mid+1, r, count*2+1);
    pushUp(count);
}
ll query(int l,int r,int count,int flag)
{
    if(num[count].l==l && num[count].r==r)
    {
        if(flag)
            return num[count].minn;
        else
            return num[count].maxx;
    }
    int mid = (num[count].l+num[count].r)/2;
    if(r<=mid)
        return query(l,r, count<<1,flag);
    else
        if(l>mid)
            return query(l, r, count<<1|1,flag);
        else
        {
            if(flag)
                return  min(query(l, mid,count<<1,flag),query(mid+1, r, count<<1|1,flag));
            else
                return  max(query(l, mid,count<<1,flag),query(mid+1, r, count<<1|1,flag));
        }
}
void update(int count,int pos,int val)
{
    if(num[count].l==num[count].r)
    {
        num[count].minn=val;
        num[count].maxx=val;
        return ;
    }
    int mid = (num[count].l+num[count].r )/2;
    if(pos <= mid)
    {
        update(count<<1,pos,val);
    }
    else
        update(count<<1|1,pos,val);
    
    pushUp(count);
}
int main()
{
    cin.sync_with_stdio(false);
    int t,n;
    cin>>t;
    while(t--)
    {
        cin>>n;
        n = pow(2,n);
        for(int i = 1; i <= n; i++)
            cin>>pp[i];
        
        built_tree(1, n, 1);
        
        int q,flag,l,r;
        cin>>q;
        for(int i = 1; i <= q; i++)
        {
            cin>>flag>>l>>r;
            if(flag==1)
            {
                ll minn = query(l+1,r+1,1,1);
                ll maxx = query(l+1,r+1,1,0);
                if(maxx<0)
                    cout<<maxx*maxx<<endl;
                else
                {
                    if(minn<0)
                        cout<<minn*maxx<<endl;
                    else
                        cout<<minn*minn<<endl;
                }
            }
            else
            {
                update(1,l+1,r);
            }
        }
    }
    return 0;
}