Educational Codeforces Round 108 (Rated for Div. 2) D. Maximum Sum of Products (区间dp)

• 题意:有两个长度为\(n\)的数组\(a\)和\(b\),你可以对\(a\)的子数组反转一次,问你\(\sum_{i=1}^{n}a_i*b_i\)的最大值.

• 题解:一眼区间dp的题目,对于长度为\(len\)的子区间\([l,r]\),它可以从\([l+1,r-1]\)转移过来,那么我们设\(dp[i][j]\)为子区间\([i,j]\)能得到的最大贡献,转移方程就为:\(dp[i][j]=dp[i+1][j-1]+a[i]*b[j]+a[j]*b[i]\).每次用当前区间的\(dp\)再加上左边和右边的前缀更新答案即可.

• 代码:

  #include <bits/stdc++.h>
  #define ll long long
  #define fi first
  #define se second
  #define pb push_back
  #define me memset
  #define rep(a,b,c) for(int a=b;a<=c;++a)
  #define per(a,b,c) for(int a=b;a>=c;--a)
  const int N = 5005 ;
  const int mod = 1e9 + 7;
  const int INF = 0x3f3f3f3f;
  using namespace std;
  typedef pair<int,int> PII;
  typedef pair<ll,ll> PLL;
  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;}
   
  int n;
  ll a[N];
  ll b[N];
  ll dp[N][N];
  ll pre[N];
   
  int main() {
      ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
  	cin>>n;
  	rep(i,1,n) cin>>a[i];
  	rep(i,1,n) cin>>b[i];
  	ll res=0;
  	rep(i,1,n){
  		res+=a[i]*b[i];
  		dp[i][i]=a[i]*b[i];
  	}
   
  	rep(i,1,n){
  		pre[i]=pre[i-1]+a[i]*b[i];
  	}
  	
  	ll ans=pre[n];
  	for(int len=1;len<n;++len){
  		for(int i=1;i+len<=n;++i){
  			int j=i+len;
  			dp[i][j]=a[i]*b[j]+a[j]*b[i]+dp[i+1][j-1];
  			ans=max(ans,pre[i-1]+dp[i][j]+pre[n]-pre[j]);
  		}
  	}
   
  	cout<<ans<<'\n';
   
      return 0;
  }