AtCoder Beginner Contest 163 **E - Active Infants** （DP）

AtCoder Beginner Contest 163 E - Active Infants （DP）

Problem Statement

There are NN children standing in a line from left to right. The activeness of the ii-th child from the left is AiAi.

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

When a child who originally occupies the xx-th position from the left in the line moves to the yy-th position from the left, that child earns Ax×|x−y|Ax×|x−y| happiness points.

Find the maximum total happiness points the children can earn.

题意：

有\(\mathit n\)个物品，第\(\mathit i\)个物品有\(a_i\)的能量，

现在你可以重新将这\(\mathit n\)个物品排列，排列的顺序可以是任意的，

第\(\mathit i\)个物品在新排列中如果被安排在第\(\mathit j\)个位置上，可以产生\(|j-i|*a_i\)的价值，

问采取最优排列的情况下，所有物品产生的总价值是多少？

思路：

首先定义\(b_i.first=a_i,b_i.second=i\)

然后按照\(b_i.first\)进行降序排序，

定义\(dp[i][j]\)代表当前放了\(i+j\)个物品，其中\(\mathit i\)个物品放在紧靠着左边，$\mathit j $个物品放在紧靠着右边时产生的最大价值。

然后转移即可。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <iomanip>
#include <sstream>
#include <bitset>
#include <unordered_map>
// #include <bits/stdc++.h>
#define ALL(x) (x).begin(), (x).end()
#define sz(a) int(a.size())
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), '\0', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define chu(x)  if(DEBUG_Switch) cout<<"["<<#x<<" "<<(x)<<"]"<<endl
#define du3(a,b,c) scanf("%d %d %d",&(a),&(b),&(c))
#define du2(a,b) scanf("%d %d",&(a),&(b))
#define du1(a) scanf("%d",&(a));
using namespace std;
typedef long long ll;
ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;}
ll lcm(ll a, ll b) {return a / gcd(a, b) * b;}
ll powmod(ll a, ll b, ll MOD) { if (a == 0ll) {return 0ll;} a %= MOD; ll ans = 1; while (b) {if (b & 1) {ans = ans * a % MOD;} a = a * a % MOD; b >>= 1;} return ans;}
ll poww(ll a, ll b) { if (a == 0ll) {return 0ll;} ll ans = 1; while (b) {if (b & 1) {ans = ans * a ;} a = a * a ; b >>= 1;} return ans;}
void Pv(const vector<int> &V) {int Len = sz(V); for (int i = 0; i < Len; ++i) {printf("%d", V[i] ); if (i != Len - 1) {printf(" ");} else {printf("\n");}}}
void Pvl(const vector<ll> &V) {int Len = sz(V); for (int i = 0; i < Len; ++i) {printf("%lld", V[i] ); if (i != Len - 1) {printf(" ");} else {printf("\n");}}}
inline long long readll() {long long tmp = 0, fh = 1; char c = getchar(); while (c < '0' || c > '9') {if (c == '-') fh = -1; c = getchar();} while (c >= '0' && c <= '9') tmp = tmp * 10 + c - 48, c = getchar(); return tmp * fh;}
inline int readint() {int tmp = 0, fh = 1; char c = getchar(); while (c < '0' || c > '9') {if (c == '-') fh = -1; c = getchar();} while (c >= '0' && c <= '9') tmp = tmp * 10 + c - 48, c = getchar(); return tmp * fh;}
void pvarr_int(int *arr, int n, int strat = 1) {if (strat == 0) {n--;} repd(i, strat, n) {printf("%d%c", arr[i], i == n ? '\n' : ' ');}}
void pvarr_LL(ll *arr, int n, int strat = 1) {if (strat == 0) {n--;} repd(i, strat, n) {printf("%lld%c", arr[i], i == n ? '\n' : ' ');}}
const int maxn = 2000 + 10;
const int inf = 0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
#define DEBUG_Switch 0
int n;
int a[maxn];
pii b[maxn];
bool cmp(pii & aa, pii & bb )
{
    return aa.fi > bb.fi;
}
ll dp[maxn][maxn];
int main()
{
#if DEBUG_Switch
    freopen("C:\\code\\input.txt", "r", stdin);
#endif
    //freopen("C:\\code\\output.txt","r",stdin);
    n = readint();
    repd(i, 1, n)
    {
        a[i] = readint();
        b[i] = mp(a[i], i);
    }
    sort(b + 1, b + 1 + n, cmp);
    repd(i, 0, n - 1)
    {
        for (int j = 0; j + i < n; ++j)
        {
            int k = i + j + 1;
            dp[i + 1][j] = max(dp[i + 1][j], dp[i][j] + 1ll * b[k].fi * abs(b[k].se - (i + 1)) );
            dp[i][j + 1] = max(dp[i][j + 1], dp[i][j] + 1ll * b[k].fi * abs(b[k].se - (n - j)) );
        }
    }
    ll ans = 0ll;
    repd(i, 0, n)
    {
        ans = max(ans, dp[i][n - i]);
    }
    printf("%lld\n", ans );
    return 0;
}