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
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 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：暴力 暴力 暴力
直接暴力寻找满足条件的前边的小于当前的面积的最小值，找后边满足大于当前面积的最小值。然后记录比较，跑完全部，就会得到最小的值。就酱。。。
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;
}