987C Three displays

C. Three displays

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

It is the middle of 2018 and Maria Stepanovna, who lives outside Krasnokamensk (a town in Zabaikalsky region), wants to rent three displays to highlight an important problem.

There are $n$ displays placed along a road, and the $i$-th of them can display a text with font size $s_i$ only. Maria Stepanovna wants to rent such three displays with indices $i<j<k$ that the font size increases if you move along the road in a particular direction. Namely, the condition $s_i<s_j<s_k$ should be held.

The rent cost is for the $i$-th display is $c_i$. Please determine the smallest cost Maria Stepanovna should pay.

Input

The first line contains a single integer $n$ ($3\le n\le 3000$) — the number of displays.

The second line contains $n$ integers $s_1,s_2,\dots ,s_n$ ($1\le s_i\le {10}^9$) — the font sizes on the displays in the order they stand along the road.

The third line contains $n$ integers $c_1,c_2,\dots ,c_n$ ($1\le c_i\le {10}^8$) — the rent costs for each display.

Output

If there are no three displays that satisfy the criteria, print -1. Otherwise print a single integer — the minimum total rent cost of three displays with indices $i<j<k$ such that $s_i<s_j<s_k$.

Examples

input

Copy

5
2 4 5 4 10
40 30 20 10 40

output

Copy

90

input

Copy

3
100 101 100
2 4 5

output

Copy

-1

input

Copy

10
1 2 3 4 5 6 7 8 9 10
10 13 11 14 15 12 13 13 18 13

output

Copy

33

Note

In the first example you can, for example, choose displays $1$, $4$ and $5$, because $s_1<s_4<s_5$($2<4<10$), and the rent cost is $40+10+40=90$.

In the second example you can't select a valid triple of indices, so the answer is -1.

题意：道路上有n个显示器，让你选择三个显示器，每个显示器都有显示字体的大小，和费用，你需要租三个显示器，每个显示器显示字体的大小，逐渐递增，然后让他们花费最小。输入n,然后下面一行每个显示器的能够显示字体的大小，然后n个显示器的花费。

题解：暴力 或者dp  暴力就先选择一个显示器，找前面尺寸比它小的，后面尺寸比他大的一个显示器。然后记录费用，取最小值。dp的话，dp学的不好，看别人博客：因为只用选出三个物品，dp[i,j]代表第ii个物品作为选出的三个物品中的第

k个物品的最小花费，显然有：dp[i,j]=min(dp[i,j−1]+c[i]，dp[i][j])。

#include<bits/stdc++.h>
using namespace std;
int main()
{
    long long n,s[3010],c[3010],ans=1e10;
    cin>>n;
    for(int i=0; i<n; i++)
        cin>>s[i];
    for(int i=0; i<n; i++)
        cin>>c[i];
    for(int i=0; i<n; i++)
    {
        long long min1=1e10,min2=1e10;
        for(int j=0; j<i; j++)
        {
            if(s[j]<s[i])
                min1=min(min1,c[j]);
        }
        for(int j=i; j<n; j++)
        {
            if(s[j]>s[i])
                min2=min(min2,c[j]);
        }
        ans=min(ans,min1+c[i]+min2);
    }
    if(ans>=1e10) puts("-1");
    else cout<<ans;
    return 0;
}

DP：

#include<bits/stdc++.h>
using namespace std;
int main()
{
    long long n,s[3010],c[3010],dp[3010][10],ans=1e10;
    cin>>n;
    for(int i=0; i<n; i++)
        cin>>s[i];
    for(int i=0; i<n; i++)
        cin>>c[i];
        memset(dp,0x3f3f3f,sizeof(dp));
    for(int i=0; i<n; i++)
    {
        dp[i][1]=c[i];
        for(int j=2; j<=3; j++)
            for(int k=0; k<i; k++)
                if(s[i]>s[k])
                    dp[i][j]=min(dp[i][j],dp[k][j-1]+c[i]);
    }
    for(int i=0; i<n; i++)
        ans=min(dp[i][3],ans);
    if(ans>=1e10) puts("-1");
    else cout<<ans;
    return 0;
}