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=10;
const int MOD=1000000007;
const int MAX=0x7f7f7f7f;
const double PI=3.14159265358979323;
const int N=150004;
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+1]);
    max1[k]=max(max1[k+k],max1[k+k+1]);
}
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)>>1;
    change(l,mid,x,y,k+k);
    change(mid+1,r,x,y,k+k+1);
    update(k);
}
void query(int l,int r,int a,int b,int k,int d[])
{
    if(r<a || b<l)
    {
        d[0]=MAX;
        d[1]=-MAX;
        return;
    }
    if(a<=l && r<=b)
    {
        d[0]=min1[k];
        d[1]=max1[k];
        return;
    }
    if(l==r)
    {
        d[0]=min1[k];
        d[1]=max1[k];
        return;
    }
    int mid=(l+r)>>1;
    int d1[2],d2[2];
    query(l,mid,a,b,k+k,d1);
    query(mid+1,r,a,b,k+k+1,d2);
    update(k);
    
    d[0]=min(d1[0],d2[0]);
    d[1]=max(d1[1],d2[1]);
}
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(2,n);
        for(i=1;i<=n;i++)
        {
            scanf("%d",&x);
            min1[n+i-1]=x;
            max1[n+i-1]=x;
        }
        for(i=n-1;i;i--)
            update(i);
        scanf("%d",&m);
        for(i=1;i<=m;i++)
        {
            scanf("%d%d%d",&z,&x,&y);
            if(z==1)
            {
                x++,y++;
                int d[2];
                query(1,n,x,y,1,d);
                if(0<=d[0])
                    aans=1LL*d[0]*d[0];
                else if(d[0]<0 && 0<=d[1])
                    aans=1LL*d[0]*d[1];
                else  aans=1LL*d[1]*d[1];
                printf("%lld
",aans);
            }
            else
            {
                x++;
                change(1,n,x,y,1);
            }
        }
    }
    return 0;
}
/*
//

//
*/