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=sumlimits_{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;
}