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=;
 int n;
 struct node
 {
     long long b,s;
 }Tree[Max<<];
 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)/;
         build(k*,l,mid);
         build(k*+,mid+,r);
         Tree[k].b=max(Tree[k*].b, Tree[k*+].b);
         Tree[k].s=min(Tree[k*].s, Tree[k*+].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)/;
         if(l <= mid)query(a,mid,k*,l,r);
         if(r > mid) query(mid+,b,k*+,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)/;
         if(pos <= mid) update(l,mid,k*,pos,v);
         if(pos > mid) update(mid+,r,k*+,pos,v);
         Tree[k].b=max(Tree[k*].b, Tree[k*+].b);
         Tree[k].s=min(Tree[k*].s, Tree[k*+].s);
     }
 }

 int main()
 {
     int T,l,r,pos;
     long long val;
     int k,c,order;
     scanf("%d",&T);
     for(int t=;t<=T;t++)
     {

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