ACM-ICPC国际大学生程序设计竞赛北京赛区(2017)网络赛 Minimum

最新推荐文章于 2018-11-08 14:33:26 发布

Jade1998

最新推荐文章于 2018-11-08 14:33:26 发布

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分类专栏：树状数组

本文链接：https://blog.youkuaiyun.com/qq_37605573/article/details/78072315

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本文介绍了一种使用树状数组或线段树解决区间查询及更新问题的方法，具体为查询指定区间内两数乘积的最小值，并允许更新指定位置的数值。

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题目9 : 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

题意：查询给定区间两个数字乘积的最小值注意数字存在负数！！！！！首先求出该区间的最小值和最大值，若最小值大于等于0，那么数字乘积的最小值就是最小值的平方；反之，如果最大值小于0，那乘积的最小值一定是正的，即为最大值的平方，否则就为最小值和最大值的乘积求区间最值问题并修改则用树状数组或线段树代码如下：

#include <iostream>  
#include <stdio.h>  
#include <stdlib.h>  
using namespace std; 
typedef long  long int ll; 
   
const int MAXN =131084;  
const int INF=0x3f3f3f3f;
ll a[MAXN], h[MAXN], l[MAXN];  
ll n,m;
ll pow1(ll a,ll b)
{
    ll res=1,base=a;
    while(b)
    {
        if(b&1)
            res*=base;
        base*=base;
        b=b>>1;
    }
    return res;
}  
   
ll lowbit(ll x)  
{  
    return x & (-x);  
}  
void updata_min(ll x)  
{  
    ll lx, i;  
    while (x <= n)  
    {  
        h[x] = a[x];  
        lx = lowbit(x);  
        for (i=1; i<lx; i<<=1)  
            h[x] = min(h[x], h[x-i]);  
        x += lowbit(x);  
    }  
} 
void  updata_max(ll x)
{
	ll lx, i;  
    while (x <= n)  
    {  
        l[x] = a[x];  
        lx = lowbit(x);  
        for (i=1; i<lx; i<<=1)  
            l[x] = max(l[x], l[x-i]);  
        x += lowbit(x);  
    }  
}
ll query_max(ll x, ll y)  
{  
    ll ans = -INF;  
    while (y >= x)  
    {  
        ans = max(a[y], ans);  
        y --;  
        for (; y-lowbit(y) >= x; y -= lowbit(y))  
            ans = max(l[y], ans);  
    }  
    return ans;  
}  
ll query_min(ll x, ll y)  
{  
    ll ans = INF;  
    while (y >= x)  
    {  
        ans = min(a[y], ans);  
        y --;  
        for (; y-lowbit(y) >= x; y -= lowbit(y))  
            ans = min(h[y], ans);  
    }  
    return ans;  
}  
int main()  
{  
    ll i, j, x, y, ans_min,ans_max;  
    char c;  
    int t; 
    scanf("%d",&t);
    while(t--)
    {
    	scanf("%lld",&n);
    	n=pow1(2,n);
    	for (i=1; i<=n; i++)  
            h[i] = 0; 
        for (i=1;i<=n;i++)
        	l[i] = 0;
    	for (i=1; i<=n; i++)  
        {  
            scanf("%lld",&a[i]);  
            updata_min(i);  
            updata_max(i);
        }  
        scanf("%d",&m);
        for (i=1; i<=m; i++)  
        {  
        	int p;
            scanf("%d%lld%lld",&p,&x,&y); 
            //cout<<p<<endl;
            //cout<<endl;
			if(p==1) 
			{
				ans_min = query_min(x+1,y+1);
				//ans_max	= query_max(x+1,y+1);
				//cout<<ans_min<<"  "<<ans_max<<endl; 
				if(ans_min>=0)
					printf("%lld\n",ans_min*ans_min);
				else
				{
					ans_max	= query_max(x+1 ,y+1);
					if(ans_max<=0)
						printf("%lld\n",ans_max*ans_max);
					else	 
            			printf("%lld\n",ans_min*ans_max);  
				}
				
			}
			else
			{
				a[x+1]=y;
				updata_min(x+1);
				updata_max(x+1);
			}
        }  
    }  
    return 0;  
}