C. Three displays

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本文介绍了一个算法问题，即如何在给定的两组序列中找到三个元素，使得这三个元素的第一个序列呈现严格递增的趋势，并且这三个元素在第二个序列中的和最小。通过从前向后和从后向前两次遍历的方法来解决这个问题。

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
inputCopy
5
2 4 5 4 10
40 30 20 10 40
outputCopy
90
inputCopy
3
100 101 100
2 4 5
outputCopy
-1
inputCopy
10
1 2 3 4 5 6 7 8 9 10
10 13 11 14 15 12 13 13 18 13
outputCopy
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：给你两个n的序列要你根据第一个序列（严格单调递增的方式）在第二个序列里找3个数加起来，输出最小的一个。

思路：先从前往后枚举两个最小的。在从后往前找一个加上使其最小，最后遍历剩下的三元组。

#include<bits/stdc++.h>
#define ll long long

using namespace std;
const int maxn=4000;
const int INF=4*1e9;
struct node{
    int num,id;
};
struct node1{
    ll h,Sum;
};
node a[maxn];
node1 sum[maxn];
int main(){
    int n;
    cin>>n;
    for(int i=1;i<=n;i++){
        sum[i].Sum=INF;
    }
    for(int i=1;i<=n;i++){
        cin>>a[i].id;
    }
    for(int i=1;i<=n;i++){
        cin>>a[i].num;
    }
    ll Min=4*1e9;
    for(int i=1;i<=n-1;i++){
              Min=4*1e9;
        for(int j=i+1;j<=n;j++){
            if(a[i].id<a[j].id){
                if(a[i].num+a[j].num<Min)
                {
                    Min=a[i].num+a[j].num;
                }
            }
        }
        sum[i].Sum=Min;
        sum[i].h=2;
    }
    //cout<<sum[2].Sum<<endl;

    for(int i=n-1;i>=2;i--){
             Min=4*1e9;
        for(int j=i-1;j>=1;j--){
            if(a[i].id>a[j].id){
                if(sum[i].Sum+a[j].num<Min){
                       Min=sum[i].Sum+a[j].num;
                }
            }
        }
        sum[i].Sum=Min;
        sum[i].h=3;
    }
    Min=4*1e9;
    int i,k;
    for( i=1;i<=n;i++){
        if(Min>sum[i].Sum&&sum[i].h==3){
            Min=sum[i].Sum;
            k=i;
        }
    }
    if(Min<INF){
    cout<<Min<<endl;
    }
    else
        cout<<"-1"<<endl;
    return 0;

}