Codeforces Round #485 (Div. 2) Three displays

最新推荐文章于 2024-07-15 17:55:50 发布

原创最新推荐文章于 2024-07-15 17:55:50 发布 · 180 阅读

CC 4.0 BY-SA版权

本文介绍了一道关于选择三个递增大小的显示屏以达到最小租赁成本的问题。通过使用动态规划方法，文章详细解释了如何有效地解决这个问题，并给出了具体的实现代码。

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 \leq n \leq 3000

) — the number of displays.

The second line contains $n$

integers

s_{1}, s_{2}, \dots, s_{n}

(

1 \leq s_{i} \leq 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 \leq c_{i} \leq 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

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

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

给你带权值的数，选三个递增的数，使权值最小
思路
dp，先找出每个数结尾两个数最小权值，再dp找第三个数

#include <iostream>
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define inf 0x3f3f3f3f
ll dp1[3100],dp2[3100];
struct node
{
    ll data,num;
}a[3100];

int main()
{
    ll n ,i ,j;
    while(cin>>n)
    {
        for(i=0;i<n;i++)
            cin>>a[i].data;
        for(i=0;i<n;i++)
            cin>>a[i].num;
        memset(dp1,0,sizeof(dp1));
        memset(dp2,0,sizeof(dp2));
        for(i=0;i<n;i++)
        {
            dp1[i] = inf;
            for(j=0;j<i;j++)
            {
                if(a[j].data<a[i].data) //
                    dp1[i] = min(dp1[i],a[i].num+a[j].num);//两个数权值最小
            }
        }
        for(i=0;i<n;i++)
        {
            dp2[i] = inf;
            for(j=0;j<i;j++)
            {
                if(a[j].data<a[i].data&&dp1[j]<inf) //被选得数必须dp第一次中符合得数
                {
                    dp2[i] = min(dp2[i],dp1[j]+a[i].num);
                }
            }
        }
        ll ans = inf;
        for(i=0;i<n;i++)
        {
            //cout<<dp1[i]<<" "<<dp2[i]<<endl;
            ans = min(ans,dp2[i]);
        }
        if(ans == inf)
            cout<<"-1"<<endl;
        else
            cout<<ans<<endl;
    }
    return 0;
}