E - Active Infants

Posted 2021-02-22 asunayi

Time Limit: 2 sec / Memory Limit: 1024 MB

 

Score : $500$points

 

ps:一道dp，这次真的是我自己写的！一直没过，把int都改成ll过了= =没注意数据范围qaq

Problem Statement

There are $\mathrm{NN}$ children standing in a line from left to right. The activeness of the $\mathrm{ii}$-th child from the left is ${\mathrm{Ai}}^{}$.

You can rearrange these children just one time in any order you like.

When a child who originally occupies the $\mathrm{xx}$-th position from the left in the line moves to the $y$-th position from the left, that child earns ${\mathrm{Ax\times |x-y|}}^{}$happiness points.

Find the maximum total happiness points the children can earn.

Constraints

• $\mathrm{2\le N\le 2000}$
• $\mathrm{1\le Ai\le 109}$
• All values in input are integers.

---

Input

Input is given from Standard Input in the following format:

$N$${\mathrm{A1}}^{}$${\mathrm{A2}}^{}$$......$${\mathrm{AN}}^{}$

Output

Print the maximum total happiness points the children can earn.

---

Sample Input 1 Copy

4
1 3 4 2

Sample Output 1 Copy

20

If we move the $1$-st child from the left to the $3$-rd position from the left, the $2$-nd child to the $4$-th position, the $3$-rd child to the $1$-st position, and the $4$-th child to the $2$-nd position, the children earns $\mathrm{1\times |1-3|+3\times |2-4|+4\times |3-1|+2\times |4-2|=20}$happiness points in total.

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Sample Input 2 Copy

Copy

6
5 5 6 1 1 1

Sample Output 2 Copy

Copy

58

---

Sample Input 3 Copy

Copy

6
8 6 9 1 2 1

Sample Output 3 Copy

Copy

85

#include <bits/stdc++.h>
using namespace std;
#define ll long long
const int mod=1e9+7;
const int N=2e3+5;
ll dp[N][N];
ll maxn;

struct nobe{
ll v,p;
}a[N];

bool cmp(nobe a,nobe b){
return a.v>b.v;
}

int main(){
ll n;
cin>>n;
for(ll i=0;i<n;i++){
    cin>>a[i].v;
    a[i].p=i+1;
}
sort(a,a+n,cmp);
for(ll i=0;i<=n;i++)
    for(ll j=0;j<=n;j++){
    if(i+j==n) {
        maxn=max(maxn,dp[i][j]);
        break;
        }
    int k=i+j;
    dp[i+1][j]=max(dp[i+1][j],dp[i][j]+abs(a[k].p-i-1)*a[k].v);
    dp[i][j+1]=max(dp[i][j+1],dp[i][j]+abs(a[k].p-n+j)*a[k].v);
}
cout<<maxn<<endl;
return 0;
}