FZU - 2302 (Optimization slope dp)

Question is intended: to a ring segment is divided into k, find how to make the sub-sub-segment of minimum square sum.

Thinking: provided DP [i] [k] denotes the i-th former minimum value is divided into k segments, readily available dp [i] [k] = min (dp [i] [j], dp [i] [k-1 ] + ([i] sum -sum [j-1]) * ([i sum] -sum [j-1])). algorithm can be written in a n ^ 4, but n is 200, is the sum observed Increasing.

Provided k1 <k2.

Order F. [K2] [k1] represents a ratio k1 to k2 i.e. excellent. So F [k2] [k1] can be written as a form of expression

dp[i][k-1]+(sum[i]-sum[k1])*(sum[i]-sum[k1])<=dp[i][k-1]+(sum[i]-sum[k2])*(sum[i]-sum[k2])（式子一）.

Simplification obtained sum [k1] * sum [k1] -sum [k2] * sum [k2] + dp [k1] [k1] -dp [k2] [k1] <= 2 * (sum [k1 ] -sum [k2]) * sum [i] (equation II).

Because the sum is then incremented for the later i + 1, i + 2 ....., k1, k2 case where a constant expression is always satisfied, then the preferred ratio should always be k2 k1. Then later we will no longer be considered a k1.

We found that if F [j] [i] <= F [k] [j], i.e., k superior than j, then j can be eliminated.

#include <algorithm>
#include <stdio.h>
#include <string.h>
#include <iostream>
using namespace std;

#define INF 0x3f3f3f3f
typedef long long ll;
long long sum[300],dp[205][205];
long long a[300],b[300];

ll getx(ll k1, ll k2,ll k)
{
	return sum[k1]*sum[k1]-sum[k2]*sum[k2]+dp[k1][k-1]-dp[k2][k-1];
}
ll gety(ll k1,ll k2)
{
	return 2*(sum[k1]-sum[k2]);
}
ll q[1005];
int main()
{
	int t;cin>>t;
	while(t--)
	{
		int n,K;cin>>n>>K;
		for(int i=1;i<=n;i++) cin>>b[i];
		long long ans=INF;
		for(int l=1;l<=n;l++)
		{
			for(int j=1;j<=n;j++) a[j]=b[((j+l-1)>n?(j+l-1-n):(j+l-1))];
			for(int j=1;j<=n;j++) sum[j]=sum[j-1]+a[j],dp[j][1]=sum[j]*sum[j];
			for(int k=2;k<=K;k++)
			{
				int head=0,tail=0;
				q[tail++]=k-1;
				for(int i=k;i<=n;i++)	//前i个分为k段
				{
					while(head<tail-1&&getx(q[head+1],q[head],k)<=sum[i]*gety(q[head+1],q[head])) head++;
					//cout<<head<<endl;
					dp[i][k]=dp[q[head]][k-1]+(sum[i]-sum[q[head]])*(sum[i]-sum[q[head]]);
					//printf("%d %d %lld\n",i,k,dp[i][k]);
					while(head<tail-1&&getx(q[tail-1],q[tail-2],k)*gety(i,q[tail-1])>=getx(i,q[tail-1],k)*gety(q[tail-1],q[tail-2])) tail--; 
					q[tail++]=i;
				/*	for(int j=0;j<i;j++)		//优化掉这层 
					{
						
						dp[i][k]=min(dp[i][k],dp[j][k-1]+(sum[i]-sum[j])*(sum[i]-sum[j]));		//i~k为一段
						//printf("%d %d %lld\n",i,k,dp[i][k]);
						
					}*/
				}
			}
			ans=min(ans,dp[n][K]);
		}
		cout<<ans<<endl;
	}
	return 0;
}