Codeforces Round #485 (Div. 2) C题求三元组（思维）

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 si<sj<sk should be held.

The rent cost is for the ci. Please determine the smallest cost Maria Stepanovna should pay.

Input

The first line contains a single integer 3≤n≤3000) — the number of displays.

The second line contains 1≤si≤109) — the font sizes on the displays in the order they stand along the road.

The third line contains 1≤ci≤108) — 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 si<sj<sk.

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 40+10+40=90.

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

 

 

题意：选三个数，要求：i<j<k 且a[i]<a[j]<a[k],要求选出来的三个数的权值最小

思路：开始总想的是贪心，二分啥啥啥的。。。结果仔细想了下，他的范围是3000， O(n^3)的时间复杂度肯定不行，O（n^2）就可以过

只要我预处理第三个数，在每个数这从后面找一个权值最小且大于它的数，以此来作为第三个数即可，后面只要枚举两个数即可

 

#include<cstdio>
#include<cmath>
#include<iostream>
#define ll long long
using namespace std;
ll a[5000],b[5000],dp[5000];
int main() {
    int n;
    scanf("%d",&n);
    for(int i=0;i<n;i++)
        scanf("%d",&a[i]);
    for(int i=0;i<n;i++)
        scanf("%d",&b[i]);
    for(int i=0;i<n;i++) 
    {
        ll mn=99999999999;
        for(int j=i+1;j<n;j++) {
            if(a[i]<a[j]) {
                mn=min(mn,b[j]);
            }
        }
        dp[i]=mn;
    }
    ll ans=999999999999;
    for(int i=0;i<n;i++)
    {
        for(int j=i+1;j<n;j++)
        {
            if(a[i]<a[j]) 
            {
                if(dp[j]!=99999999999)
                    ans=min(ans,b[i]+b[j]+dp[j]);
            }
        }
    }
    if(ans==999999999999) printf("-1\n");
    else cout<<ans<<endl;
}

总结：总的来说这次cf div2的题目不是很难，只是自己刷提还是刷的太少了，没想到思路就卡到了，训练太少，刷题太慢，需要好好反省

规律题总结，看到那种遍历一遍就超时那就肯定是规律题，一 班自己先把公式推出来，然后写个小枚举，把答案输出出来，然后自己再把递推公式推出来，再求解即可