PQF的博客

This is the second problem of Div.3. But I think it is not easy.

Problem:1029B

Simple Meaning of Problem:

There is a sequence $a$ and you should choose the longest sequence $b$ { $a_{i_1}, a_{i_2}, \dots, a_{i_p}$ } ( $p$ is the length of $b$ ) such that:each $j∈[1,k)$ $a_{i_{j+1}} \le a_{i_j} \cdot 2$ should hold.

print the maximum number of $p$ (the length of $b$ )

First, you can enumerate which ones to choose. You will get a algorithm with $O(2^n)$ . And you will get a Time Limit Exceed.(I didn't write it, because it is valueless.)

Then, dp is another useful algorithm. So we can try to use dp.

$f[i]$ means that if you choose the number $a_i$ , you can get the maximum $p$ (only of $1$ to $i$ ).

It is very easy to get the state transition equation:

$f[i]=max{f[j]}+1$ , $j$ satisfy that $a_i \le a_j \cdot 2$ and $j \le i$ .

And the answer is $max$ { $f[i]$ } ( $i$ is $1$ to $n$ ).

code:42093398

#pragma comment (linker,"/STACK:1024000000") 
#pragma GCC optimize(2)
#pragma GCC target(arch=corie7-acx)
#include<bits/stdc++.h>
using namespace std;
int n,a[200001],f[200001];
deque <int> q;
int t,ans;
int main()
{
    scanf("%d",&n);
    for (int i=1;i<=n;i++)
    scanf("%d",&a[i]);
    f[1]=1;
    for (int i=2;i<=n;i++)
    {
        for (int j=1;j<=i-1;j++)
        if (a[j]*2>=a[i]) f[i]=max(f[i],f[j]);
        f[i]++;
    }
    for (int i=1;i<=n;i++)
    ans=max(ans,f[i]);
    printf("%d",ans);
    return 0;
}

And this code does better:42095786(In fact, it also Time Limit Exceed on Test 4)

#pragma comment (linker,"/STACK:1024000000") 
#pragma GCC optimize(2)
#pragma GCC target(arch=corie7-acx)
#include<bits/stdc++.h>
using namespace std;
int n,a[200001],f[200001];
deque <int> q;
int t,ans;
int main()
{
    scanf("%d",&n);
    for (int i=1;i<=n;i++)
    scanf("%d",&a[i]);
    f[1]=1;
    for (int i=2;i<=n;i++)
    {
        for (int j=i-1;j>=1;j--)
        if (a[j]*2>=a[i]) f[i]=max(f[i],f[j]);
            else break;
        f[i]++;
    }
    for (int i=1;i<=n;i++)
    ans=max(ans,f[i]);
    printf("%d",ans);
    return 0;
}

But, unfortunately, these two are both Time Limit Exceed. (Actually, it is inevitable.)

Because when $a_n \le a_1 \cdot 2$ , the time complexity of the code will come to $O(n^2)$ (break is useless).

So it is inevitable that the code will TLE.

Now, how can we optimize our algorithm?

In the problem, notice that the numbers are increasing.

So when $i$ is increasing, $a_i$ is increasing, too. And $\frac {a_i} {2}$ is increasing, so the number which is less than $\frac {a_i} {2}$ is also increasing.

Because of this, for the increasing number $i$ , the $j$ which is satisfy $a_i \le a_j \cdot 2$ is increasing as well.

So we can maintain a queue. The number of the queue means the number of $a$ 's number.

When we want to calculate $f[i]$ , we check the numbers of queue from the front to the back. For each $x$ , if $a_x \cdot 2 < a_i$ , we can pop $x$ . (Because for the after $j$ , $a_j \ge a_i > a_x \cdot 2$ ).

We can find the first $x$ which is satisfy $a_x \cdot 2 \ge a_i$ . Then we use $f[x]$ to calculate $f[i]$ . (It will be proven after).

Finally, we get the back of $q$ $y$ . If $f[y] \le f[i]$ , we can pop $y$ . That's because $y$ must be popped earlier than $i$ , and $f[i]$ is greater than $f[y]$ . So the following number choose $f[i]$ is greater than $f[y]$ .

And that's why the first number of the queue $x$ is the best choice to transfer $f[i]$ after popping. (In fact, it is a priority queue.)

When we code it, we can use double-end-queue (deque).

code:42036114

#pragma comment (linker,"/STACK:1024000000") 
#pragma GCC optimize(2)
#pragma GCC target(arch=corie7-acx)
#include<bits/stdc++.h>
using namespace std;
int n,a[200001],f[200001];
deque <int> q;
int t,ans;
int main()
{
    scanf("%d",&n);
    for (int i=1;i<=n;i++)
    scanf("%d",&a[i]);
    f[1]=1;
    q.push_back(1);
    for (int i=2;i<=n;i++)
    {
        while (!q.empty())
        {
            t=q.front();
            if (a[t]*2<a[i]) q.pop_front();
                else break;
        }
        if (q.empty()) t=0;
        f[i]=f[t]+1;
        while (!q.empty())
        {
            t=q.back();
            if (f[t]<f[i]) q.pop_back();
                else break;
        }
        q.push_back(i);
    }
    for (int i=1;i<=n;i++)
    ans=max(ans,f[i]);
    printf("%d",ans);
    return 0;
}

Thank you for reading my first entry. See you next time!

postscripts:

It is the first time of mine to write the solution (entry). So if there is any mistake, please specify PQF. Thanks.
Actually, I'm a Chinese. So please forgive my English level. Thanks.
Oh, writing an entry is so tiring!

其实本来是发布在CF上的，但觉得在ZROI上发一下也没啥事吧~~~

博客

Solution of 1029B. Creating new contest

Day1