#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 2k integers, a0, a1, …, a2^k-1 (-2^k ≤ ai < 2^k).
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:一共有2^k个数(0~2^k -1)两个操作
1 x y 求 区间[x,y]之间 乘积的最小值
2 x y 将ax变成y(ax=y)
由于ai的范围(-2^k ≤ ai < 2^k),存在负数,所以存在三种情况:
1 [x,y]之间最小值大于0,那么乘积的最小值为 最小值*最小值;
2 [x,y]之间最小值小于0,最大值大于0,那么最小值为 最小值*最大值;
3 [x,y]之间最小值小于0,最大值小于0,那么最小值为 最大值*最大值。
代码:
//线段树单点修改,求区间最小值
#include <iostream>
#include <cstdio>
#define MIN 0x3f3f3f3f
typedef long long LL;
using namespace std;
int n, m;
int mi[525000];
int ma[525000];
int power(int a,int n)
{
int ans=1;
while(n)
{
if(n%2)ans=ans*a;
n/=2;
a*=a;
}
return ans;
}
void pushup(int k)//自下向上更新
{
mi[k] = min(mi[k << 1], mi[k << 1 | 1 ]);
ma[k] = max(ma[k << 1], ma[k << 1 | 1 ]);
}
void build(int k, int l, int r)//初始建树,k表示当前结点的编号,l,r表示当前结点所代表的区间
{
if(l == r) { scanf("%d",&mi[k]); ma[k]=mi[k];return; }
int mid = l + r >> 1;
build(k << 1, l, mid);//构造左子树
build(k << 1 | 1, mid + 1, r);//构造右子树
pushup(k);
}
void change(int k, int l, int r, int x, int v)//单点修改,将结点x的值修改为v
{
if(l == r && l == x ) { mi[k]=ma[k] = v; return; }//当前结点为对应的叶子节点
int mid = l + r >> 1;
if(x <= mid) change(k << 1, l, mid, x, v);//修改左子区间
else change(k << 1 | 1, mid + 1, r, x, v);//修改右子区间
pushup(k);//更新相关的值
}
int query_min(int k, int l, int r, int x, int y)//询问区间最小值,k表示当前结点的编号,l,r表示当前区间,x,y表示询问区间
{
if(l >= x && r <= y) return mi[k];//询问区间包含当前区间,返回维护好的最小值
int mid = l + r >> 1;
int minn = MIN;
if(x <= mid) minn = min(minn, query_min(k << 1, l, mid, x, y));
if(y > mid) minn = min(minn, query_min(k << 1 | 1, mid + 1, r, x, y));
return minn;//否则分别处理左子区间和右子区间,用两个if忽略掉询问区间与当前区间无交集的情况
}
int query_max(int k, int l, int r, int x, int y)//询问区间最大值,k表示当前结点的编号,l,r表示当前区间,x,y表示询问区间
{
if(l >= x && r <= y) return ma[k];//询问区间包含当前区间,返回维护好的最大值
int mid = l + r >> 1;
int maxn = -MIN;
if(x <= mid) maxn = max(maxn, query_max(k << 1, l, mid, x, y));
if(y > mid) maxn = max(maxn, query_max(k << 1 | 1, mid + 1, r, x, y));
return maxn;//否则分别处理左子区间和右子区间,用两个if忽略掉询问区间与当前区间无交集的情况
}
int main()
{
int t;scanf("%d",&t);
while(t--)
{
int k;
scanf("%d",&k);
k=power(2,k);
//printf("k=%d\n",k);
build(1,0,k-1);
int q,op,a,b;
scanf("%d",&q);
for(int i=0;i<q;i++)
{
scanf("%d%d%d",&op,&a,&b);
if(op==1)
{
LL ans,minn,maxx;
minn = (LL)query_min(1,0,k-1,a,b);
maxx = (LL)query_max(1,0,k-1,a,b);
//printf("mx = %lld ,mn = %lld\n",maxx,minn);
if(minn>=0)
{
ans = minn*minn;
}
else if(minn<0&&maxx>0)ans = minn*maxx;
else ans = maxx*maxx;
printf("%lld\n", ans);//p==1表示询问区间[a,b]最小值
}
else{
change(1, 0, k-1, a, b);//p==2表示将结点a的值修改为b
}
}
}
return 0;
}