2017ICPC北京赛区网络赛 Minimum（数学+线段树）

描述

You are given a list of integers a₀, a₁, …, a_2^k-1.

You need to support two types of queries:

1. Output Min_x,y∈[l,r] {a_x∙a_y}.

2. Let a_x=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 2^k integers, a₀, a₁, …, a_2^k-1 (-2^k ≤ a_i < 2^k).

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

1. 1 l r: Output Min_x,y∈[l,r]{a_x∙a_y}. (0 ≤ l ≤ r < 2^k)

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

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

分析得出，一区间的乘积最小值，只有三种情况
1.最大值的平方（最大值为负数）
2.最小值的平方（最小值为正数）
3.最大值与最小值的乘积（最大值是正，最小值为负）
得到此结果后，只需改一下线段树的模板，将每段区间的最大最小值保存下来即可。

#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
using namespace std;
const long long Max=140000;
int n;
struct node
{
    long long b,s;
}Tree[Max<<2];
long long bb,ss;
void build(int k,int l,int r)//建线段树，k表示子节点坐标
{
    if(l == r) 
    {
        scanf("%lld",&Tree[k].b);
        Tree[k].s=Tree[k].b;
    }
    else
    {
        int mid = (l+r)/2;
        build(k*2,l,mid);
        build(k*2+1,mid+1,r);
        Tree[k].b=max(Tree[k*2].b, Tree[k*2+1].b);
        Tree[k].s=min(Tree[k*2].s, Tree[k*2+1].s);
    }
}
void query(int a,int b,int k,int l,int r)//a,b是当前查询区间，k是当前的根节点，l,r是要求查询区间
{
    if(a >= l && b <= r) 
    {
        bb=max(bb,Tree[k].b);
        ss=min(ss,Tree[k].s);
        //return min(min(Tree[k].b*Tree[k].b,Tree[k].b*Tree[k].s),Tree[k].s*Tree[k].s);
        return;
    }
    else
    {
        //long long ans = Max*Max;
        int mid = (a+b)/2;
        if(l <= mid)query(a,mid,k*2,l,r);
        if(r > mid) query(mid+1,b,k*2+1,l,r);
        return;
        //return ans;
    }
}
void update(int l,int r,int k,int pos,long long v)//l,r是查询区间，k是当前根节点，pos是查询位置
{
    if(l == r)
    {
        Tree[k].b=v;
        Tree[k].s=v;
    }
    else{
        int mid = (l+r)/2;
        if(pos <= mid) update(l,mid,k*2,pos,v);
        if(pos > mid) update(mid+1,r,k*2+1,pos,v);
        Tree[k].b=max(Tree[k*2].b, Tree[k*2+1].b);
        Tree[k].s=min(Tree[k*2].s, Tree[k*2+1].s);
    }
}

int main()
{
    int T,l,r,pos;
    long long val;
    int k,c,order;
    scanf("%d",&T);
    for(int t=1;t<=T;t++)
    {
        
        scanf("%d",&k);
        n=pow(2.0,k);
        build(1,1,n);
        scanf("%d",&c);
        while(c--)
        {
            scanf("%d",&order);
            if(order==1)
            {
                scanf("%d%d",&l,&r);
                bb=-Max;ss=Max;
                long long tmpans;
                query(1,n,1,l+1,r+1);
                tmpans= min(min(bb*bb,bb*ss),ss*ss);
                printf("%lld
",tmpans);
            }
            else if(order==2)//点更新
            {
                scanf("%d%lld",&pos,&val);
                update(1,n,1,pos+1,val);
            }
        }
    }
    return 0;
}