NOI2009 诗人小G

显然有\(f[i]=min(f[j]+w(i,j)),w(i,j)=|sum[i]-sum[j]-L-1+i-j|^P\)

\(sum_i=\sum\limits_{j=1}^ilen_j\)

通过打表，我们可以发现\(w\)满足四边形不等式。

于是我们用单调队列来维护决策点。

二分找出最优位置的分界点就可以了

/*
@Date    : 2019-08-16 20:23:22
@Author  : Adscn (adscn@qq.com)
@Link    : https://www.cnblogs.com/LLCSBlog
*/
#include<bits/stdc++.h>
using namespace std;
#define IL inline
#define RG register
#define gi getint()
#define gc getchar()
#define File(a) freopen(a".in","r",stdin);freopen(a".out","w",stdout)
IL int getint()
{
	RG int xi=0;
	RG char ch=gc;
	bool f=0;
	while(ch<'0'||ch>'9')ch=='-'?f=1:f,ch=gc;
	while(ch>='0'&&ch<='9')xi=(xi<<1)+(xi<<3)+ch-48,ch=gc;
	return f?-xi:xi;
}
template<typename T>
IL void pi(T k,char ch=0)
{
	if(k<0)k=-k,putchar('-');
	if(k>=10)pi(k/10,0);
	putchar(k%10+'0');
	if(ch)putchar(ch);
}
typedef long double ld;
typedef long long ll;
const int N=5e5+7;
ld dp[N];
long long L,P,sum[N];
int ans[N],n,T;
int Q[N];
int a[N];
inline ld fpow(ld a,ll b){
	ld t=1;
	while(b){
		if(b&1)t*=a;
		a*=a;b>>=1;
	}
	return t;
}
inline ld calc(int i,int x){
	return dp[i]+fpow(abs(sum[x]-sum[i]+x-i-L-1),P);
}
inline int hget(int a,int b){
	if(calc(a,n)<calc(b,n))return n+1;
	int l=b,r=n,ans=-1;
	while(l<=r)
	{
		int mid=(l+r)>>1;
		if(calc(b,mid)<=calc(a,mid))ans=mid,r=mid-1;
		else l=mid+1;
	}
	return ans;
}
int stk[N],top;
string s[N];
int main(void)
{
	T=gi;
	while(T--)
	{
		n=gi,L=gi,P=gi;
		for(int i=1;i<=n;++i)
		{
			cin>>s[i];
			a[i]=s[i].size();
			sum[i]=sum[i-1]+a[i];
			dp[i]=1e19;
		}
		memset(ans,0,sizeof ans);
		int l=1,r=1;
		for(int i=1;i<=n;++i)
		{
			while(l<r&&hget(Q[l],Q[l+1])<=i)++l;
			ans[i]=Q[l];
			dp[i]=calc(Q[l],i);
			while(l<r&&hget(Q[r-1],Q[r])>=hget(Q[r],i))--r;
			Q[++r]=i;
		}
		if(dp[n]>1e18)puts("Too hard to arrange");
		else{
			cout<<(ll)dp[n]<<endl;
			top=0;
			for(int tmp=n;tmp;tmp=ans[tmp])stk[++top]=tmp;
			stk[++top]=0;
			for(int i=top;i>1;--i)
			{
				for(int j=stk[i]+1;j<stk[i-1];++j)cout<<s[j]<<" ";
				cout<<s[stk[i-1]]<<endl;
			}
		}
		puts("--------------------");
	}
	return 0;
}