SXCPC2018 nucoj2007 和Mengjiji一起攻克难关

problem

#include <algorithm>
#include <iostream>
#include <cstdio>
using namespace std;
typedef long long ll;
int n, a[2005];
ll dp[2005][2005];
int main(){
	cin>>n;
	for(int i=1; i<=n; i++)
		scanf("%d", &a[i]);
	dp[2][0] = abs(a[2]-a[1]);
	for(int i=3; i<=n; i++)
		for(int j=0; j<i; j++){
			if(j==i-1){
				dp[i][j] = dp[i-1][0];
				for(int k=1; k<=i-2; k++)
					dp[i][j] = min(dp[i][j], dp[j][k]+abs(a[k]-a[i]));
			}
			else	dp[i][j] = dp[i-1][j] + abs(a[i] - a[i-1]);
		}
	ll ans=0x3f3f3f3f3f3f3f3f;
	for(int i=0; i<n; i++)
		ans = min(ans, dp[n][i]);
	cout<<ans<<endl;
	return 0;
}