C. Three displays codeforces round 485 (Div. 2)

Original link ： ​ ​https://codeforces.com/problemset/problem/987/C​​

The test sample ：

The sample input 1
5
2 4 5 4 10
40 30 20 10 40
Sample output 1
90
The sample input 2
3
100 101 100
2 4 5
Sample output 2
-1
The sample input 3
10
1 2 3 4 5 6 7 8 9 10
10 13 11 14 15 12 13 13 18 13
Sample output 3
33

Tips ：

No violence .

Their thinking ： The seniors warned against violence , Or two violent attempts . After the game, I found it was ,dp Just solve it . Okay , Let's talk about ideas , This problem we are asking to solve can make The minimum value of , among , So we can discuss one part of it first, that is , We're going to use dp To solve the , use Express satisfaction Of The minimum value of , We don't care Value , We just have to find the minimum , meanwhile Satisfy Just go . So after we get this minimum, we can get another minimum ？ Let's split this comparison into two , So our original three levels of violence for The cycle is broken down into two ,OK, Let's look at the code .

AC Code ：

Log in and copy

      
      

       
       /*
       
       

       
       

       
       *
       
       

       
       */
       
       

       
       #include<bits/stdc++.h> 
       
       //POJ I won't support it 
       
       

       
       

       
       #define rep(i,a,n) for (int i=a;i<=n;i++)
       
       //i Is a cyclic variable ,a For the initial value ,n Is the limit value , Increasing 
       
       

       
       #define per(i,a,n) for (int i=a;i>=n;i--)
       
       //i Is a cyclic variable , a For the initial value ,n Is the limit value , Decline .
       
       

       
       #define pb push_back
       
       

       
       #define IOS ios::sync_with_stdio(false);cin.tie(0); cout.tie(0)
       
       

       
       #define fi first
       
       

       
       #define se second
       
       

       
       #define mp make_pair
       
       

       
       

       
       using 
       
       namespace 
       
       std;
       
       

       
       

       
       const 
       
       int 
       
       inf 
       
       = 
       
       0x3f3f3f3f;
       
       // infinity 
       
       

       
       const 
       
       int 
       
       maxn 
       
       = 
       
       1e5;
       
       // Maximum .
       
       

       
       typedef 
       
       long 
       
       long 
       
       ll;
       
       

       
       typedef 
       
       long 
       
       double 
       
       ld;
       
       

       
       typedef 
       
       pair
       
       <
       
       ll, 
       
       ll
       
       >  
       
       pll;
       
       

       
       typedef 
       
       pair
       
       <
       
       int, 
       
       int
       
       > 
       
       pii;
       
       

       
       //******************************* Split line , The above is a custom code template ***************************************//
       
       

       
       

       
       int 
       
       n;
       
       

       
       int 
       
       a[
       
       maxn],
       
       b[
       
       maxn],
       
       dp[
       
       maxn];
       
       //dp[i] It means a[i]>a[j] bring b[i]+b[j] The minimum value , among 1<=j<i.
       
       

       
       int 
       
       main(){
       
       
  
       
       //freopen("in.txt", "r", stdin);// When submitting, you should comment out 
       
       
  
       
       IOS;
       
       
  
       
       while(
       
       cin
       
       >>
       
       n){
       
       
    
       
       rep(
       
       i,
       
       1,
       
       n)
       
       cin
       
       >>
       
       a[
       
       i];
       
       
    
       
       rep(
       
       i,
       
       1,
       
       n)
       
       cin
       
       >>
       
       b[
       
       i];
       
       
    
       
       memset(
       
       dp,
       
       inf,
       
       sizeof(
       
       dp));
       
       
    
       
       rep(
       
       i,
       
       2,
       
       n){
       
       
      
       
       rep(
       
       j,
       
       1,
       
       i
       
       -
       
       1){
       
       
        
       
       if(
       
       a[
       
       i]
       
       >
       
       a[
       
       j]){
       
       
          
       
       dp[
       
       i]
       
       =
       
       min(
       
       dp[
       
       i],
       
       b[
       
       i]
       
       +
       
       b[
       
       j]);
       
       
        }
       
       
      }
       
       
    }
       
       
    
       
       // We get it again b[i]+dp[j] The minimum value of , This is the answer we want .
       
       
    
       
       int 
       
       ans
       
       =
       
       inf;
       
       
    
       
       rep(
       
       i,
       
       2,
       
       n){
       
       
      
       
       rep(
       
       j,
       
       1,
       
       i
       
       -
       
       1){
       
       
        
       
       if(
       
       a[
       
       i]
       
       >
       
       a[
       
       j]){
       
       
          
       
       ans
       
       =
       
       min(
       
       ans,
       
       b[
       
       i]
       
       +
       
       dp[
       
       j]);
       
       
        }
       
       
      }
       
       
    }
       
       
    
       
       if(
       
       ans
       
       ==
       
       inf)
       
       
      
       
       cout
       
       <<-
       
       1
       
       <<
       
       endl;
       
       
    
       
       else
       
       
      
       
       cout
       
       <<
       
       ans
       
       <<
       
       endl;
       
       
  }
       
       
  
       
       return 
       
       0;
       
       
}
      
      
      
      

       
       
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