#1586 : Minimum（线段树）

#1586 : Minimum

时间限制:1000ms

单点时限:1000ms

内存限制:256MB

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

EmacsNormalVim

思路：输出x，y范围内的min(ax,ay)，就是找到x--y中最小值pmin和pmax，

如果最小值>=0，最小的结果一定是pmin*pmin;

如果最小值<0,就有两种情况

（1）pmax>=0，最小结果是pmin*pmax，是负数

（2）pmax<0,最小的结果是pmax*pmax，是正数。

参考文章：https://blog.csdn.net/qq_38538733/article/details/78072262

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
const int INF = 99999999;
const int maxn = 5002000;
typedef long long LL;
LL pmax,pmin;
struct Node{
	int l,r;
	LL mi,mx;
};
Node t[maxn];
void pushup(int k)
{
	t[k].mi=min(t[k*2].mi,t[k*2+1].mi);
	t[k].mx=max(t[k*2].mx,t[k*2+1].mx);
}
void build(int k,int l,int r)
{
	t[k].l=l;
	t[k].r=r;
	if(l==r)
	{
		scanf("%lld",&t[k].mi);
		t[k].mx=t[k].mi;
		return ;
	}
	int mid=(l+r)/2;
	build(k*2,l,mid);
	build(k*2+1,mid+1,r);
	pushup(k);
}
void update(int pos,int k,int val)
{
	if(t[k].l==t[k].r)
	{
		t[k].mi=val;
		t[k].mx=val;
		return ;
	}
	int mid=(t[k].l+t[k].r)/2;
	if(pos<=mid) update(pos,k*2,val);
	else update(pos,k*2+1,val);
	pushup(k);
}
void query(int k,int l,int r)
{
	if(t[k].l==l&&t[k].r==r)
	{
		pmin=min(pmin,t[k].mi);
		pmax=max(pmax,t[k].mx);
		return ;
	}
	int mid=(t[k].l+t[k].r)/2;
	if(r<=mid) query(k*2,l,r);
	else if(l>mid) query(k*2+1,l,r);
	else
	{
		query(k*2,l,mid);
		query(k*2+1,mid+1,r);
	}
}
int main(void)
{
	int n,m,k,i,j,x,y,z,q,T;
	scanf("%d",&T);
	while(T--)
	{
		scanf("%d",&k);
		n=1;
		for(i=0;i<k;i++) n*=2;
		build(1,0,n-1);
		scanf("%d",&q);
		while(q--)
		{
			scanf("%d%d%d",&z,&x,&y);
			if(z==1)
			{
				pmin=INF;
				pmax=-INF;
				query(1,x,y);
				if(pmin>=0) printf("%lld\n",pmin*pmin);
				else
				{
					if(pmax>=0) printf("%lld\n",pmax*pmin);
					else printf("%lld\n",pmax*pmax);
				}
			}
			else if(z==2)
			{
				update(x,1,y);
			}
		}
	}
	return 0;
}