hihoCoder 1586 Minimum 【线段树】 (ACM-ICPC国际大学生程序设计竞赛北京赛区(2017)网络赛)

#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 2^k integers, a0, a1, …, a2^k-1 (-2^k ≤ ai < 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 Minx,y∈[l,r]{ax∙ay}. (0 ≤ l ≤ r < 2^k)

2. 2 x y: Let ax=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

题目链接：

　　http://hihocoder.com/problemset/problem/1586

题目大意：

　　给2ⁿ个数，两种操作

　　1 输出Min_x,y∈[l,r] {a_x*a_y}.

　　2 令a_x=y

题目思路：

　　【线段树】

　　记录每个区间的最大最小值，并维护，询问时获取当前区间最大最小值，分情况讨论

　　0在最小值左边，0在最小值和最大值之间，0在最大值右边，共3种情况。

　　分别计算答案即可。

　　修改时修改单点值。

 /****************************************************

     Author : Coolxxx
     Copyright 2017 by Coolxxx. All rights reserved.
     BLOG : http://blog.csdn.net/u010568270

 ****************************************************/
 #include<bits/stdc++.h>
 #pragma comment(linker,"/STACK:1024000000,1024000000")
 #define abs(a) ((a)>0?(a):(-(a)))
 #define lowbit(a) (a&(-a))
 #define sqr(a) ((a)*(a))
 #define mem(a,b) memset(a,b,sizeof(a))
 const double EPS=0.00001;
 const int J=;
 const int MOD=;
 const int MAX=0x7f7f7f7f;
 const double PI=3.14159265358979323;
 const int N=;
 using namespace std;
 typedef long long LL;
 double anss;
 LL aans;
 int cas,cass;
 int n,m,lll,ans;
 int min1[N+N],max1[N+N];
 void update(int k)
 {
     min1[k]=min(min1[k+k],min1[k+k+]);
     max1[k]=max(max1[k+k],max1[k+k+]);
 }
 void change(int l,int r,int x,int y,int k)
 {
     if(r<x || x<l)return;
     if(l==r)
     {
         min1[k]=y;
         max1[k]=y;
         return;
     }
     int mid=(l+r)>>;
     change(l,mid,x,y,k+k);
     change(mid+,r,x,y,k+k+);
     update(k);
 }
 void query(int l,int r,int a,int b,int k,int d[])
 {
     if(r<a || b<l)
     {
         d[]=MAX;
         d[]=-MAX;
         return;
     }
     if(a<=l && r<=b)
     {
         d[]=min1[k];
         d[]=max1[k];
         return;
     }
     if(l==r)
     {
         d[]=min1[k];
         d[]=max1[k];
         return;
     }
     int mid=(l+r)>>;
     int d1[],d2[];
     query(l,mid,a,b,k+k,d1);
     query(mid+,r,a,b,k+k+,d2);
     update(k);

     d[]=min(d1[],d2[]);
     d[]=max(d1[],d2[]);
 }
 int main()
 {
     #ifndef ONLINE_JUDGE
     freopen("1.txt","r",stdin);
 //    freopen("2.txt","w",stdout);
     #endif
     int i,j,k;
     int x,y,z;
     for(scanf("%d",&cass);cass;cass--)
 //    init();
 //    for(scanf("%d",&cas),cass=1;cass<=cas;cass++)
 //    while(~scanf("%d",&n))
     {
         scanf("%d",&n);
         n=pow(,n);
         for(i=;i<=n;i++)
         {
             scanf("%d",&x);
             min1[n+i-]=x;
             max1[n+i-]=x;
         }
         for(i=n-;i;i--)
             update(i);
         scanf("%d",&m);
         for(i=;i<=m;i++)
         {
             scanf("%d%d%d",&z,&x,&y);
             if(z==)
             {
                 x++,y++;
                 int d[];
                 query(,n,x,y,,d);
                 if(<=d[])
                     aans=1LL*d[]*d[];
                 else if(d[]< && <=d[])
                     aans=1LL*d[]*d[];
                 else  aans=1LL*d[]*d[];
                 printf("%lld\n",aans);
             }
             else
             {
                 x++;
                 change(,n,x,y,);
             }
         }
     }
     return ;
 }
 /*
 //

 //
 */