题目：
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 nn displays placed along a road, and the ii-th of them can display a text with font size sisi only. Maria Stepanovna wants to rent such three displays with indices i<j<ki<j<k that the font size increases if you move along the road in a particular direction. Namely, the condition si<sj<sksi<sj<sk should be held.
The rent cost is for the ii-th display is cici. Please determine the smallest cost Maria Stepanovna should pay.
Input
The first line contains a single integer nn (3≤n≤30003≤n≤3000) — the number of displays.
The second line contains nn integers s1,s2,…,sns1,s2,…,sn (1≤si≤1091≤si≤109) — the font sizes on the displays in the order they stand along the road.
The third line contains nn integers c1,c2,…,cnc1,c2,…,cn (1≤ci≤1081≤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 i<j<ki<j<k such that si<sj<sksi<sj<sk.
Examples
Input
5 2 4 5 4 10 40 30 20 10 40
Output
90
Input
3 100 101 100 2 4 5
Output
-1
Input
10 1 2 3 4 5 6 7 8 9 10 10 13 11 14 15 12 13 13 18 13
Output
33
Note
In the first example you can, for example, choose displays 11, 44 and 55, because s1<s4<s5s1<s4<s5 (2<4<102<4<10), and the rent cost is 40+10+40=9040+10+40=90.
In the second example you can't select a valid triple of indices, so the answer is -1.
解题思路：就是转化一下思维，先确定中间位置的，再去找前后最小的存在。简单的暴力，做的时候还在担心会不会超时。
ac代码：
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define maxn 100000
#define inf 1e9
using namespace std;
int n;
int s[maxn],c[maxn];
int main()
{
while(scanf("%d",&n)!=EOF)
{
int x,y,z,ans=inf;
for(int i=0;i<n;i++)
scanf("%d",&s[i]);
for(int i=0;i<n;i++)
scanf("%d",&c[i]);
for(int i=0;i<n;i++)
{
y=c[i];
x=z=inf;
for(int j=0;j<i;j++)
{
if(s[j]<s[i])
x=min(x,c[j]);
}
for(int j=i+1;j<n;j++)
{
if(s[j]>s[i])
z=min(z,c[j]);
}
ans=min(ans,x+y+z);
}
if(ans>=inf) printf("-1\n");
else
printf("%d\n",ans);
}
return 0;
}做到这道暴力题，忽然想起来，qduoj热身有一道很有意思的暴力题，也挂出来吧
博客链接：https://blog.csdn.net/qq_42505741/article/details/81584658