"Slope" Toy packing

Test instructions

There are n toys to be boxed, the length of each toy is c[i], stipulated in the packing, must be in strict accordance with the order given, and any two toys in the same box must and only one unit length, in other words, if you want to load a box numbered I~j toys, The length of the box must be only l=j-i+sigma[c[k]], the cost of each box with a length of L is P = (l-l) 2, where l is a given constant. You are now required to pack all your toys at the least cost, and the number of boxes does not matter.

Ideas:

Dp[i]=min (dp[j]+ (sum[i]-sum[j]+i-j-1-l) ^2) (J<i)

Make F[i]=sum[i]+i,c=1+l

Then Dp[i]=min (dp[j]+ (f[i]-f[j]-c) ^2)

1. Proving monotonicity of decisions

K-decision-making and J-decision based on state I, i.e.

dp[k]+ (f[i]-f[k]-c) ^2<=dp[j]+ (f[i]-dp[j]-c) ^2

For a post-state t,dp[t]=dp[i]+v;

To prove dp[k]+ (f[t]-f[k]-c) ^2<=dp[j]+ (f[t]-f[j]-c) ^2

As long as the card

dp[k]+ (f[i]+v-f[k]-c) ^2<=dp[j]+ (f[i]+v-f[j]-c) ^2

As long as the card

dp[k]+ (f[i]-f[k]-c) ^2+2*v* (f[i]-f[k]-c) +v^2<=dp[j]+ (f[i]-f[j]-c) ^2+2*v* (f[i]-f[j]-c) +v^2

As long as the card

2*v* (f[i]-f[k]-c) <=2*v* (f[i]-f[j]-c)

That is f[k]>=f[j] (obviously)

Proof complete.

2. Calculate the slope equation

Because dp[k]+ (f[i]-f[k]-c) ^2<=dp[j]+ (f[i]-f[j]-c) ^2

Expand

dp[k]+f[i]^2-2*f[i]* (f[k]+c) + (f[k]+c) ^2<=dp[j]+f[i]^2-2*f[i]* (f[j]+c) + (f[j]+c) ^2

That

dp[k]-2*f[i]* (f[k]+c) + (f[k]+c) ^2<=dp[j]-2*f[i]* (f[j]+c) + (f[j]+c) ^2

i.e. (dp[k]+ (f[k]+c) ^2-dp[j]-(f[j]+c) ^2)/2* (F[k]-f[j]) <=f[i]

F[i] is monotonically increasing, we use the queue to maintain a lower convex hull, each time we take out the team head as a decision

When I join decision I, the tail of the team is q[r], the previous one is q[r-1]

When the slope (q[r],i) < slope (Q[r-1],q[r]) is met, it is clear that the end of the team is not valid and will eject (turn from)

Code:

1#include <map>2#include <Set>3#include <ctime>4#include <cstdio>5#include <cstring>6#include <vector>7#include <cstdlib>8#include <iostream>9#include <algorithm>Ten #defineINF 1000000000 One #definell Long Long A using namespacestd; - ll Read () - { thell x=0, f=1;CharCh=GetChar (); -      while(ch<'0'|| Ch>'9'){if(ch=='-') f=-1; ch=GetChar ();} -      while(ch>='0'&&ch<='9') {x=x*Ten+ch-'0'; ch=GetChar ();} -     returnx*F; + } - intN,l,l,r; + intc[50005],q[50005]; All s[50005],f[50005],c; at DoubleSlopintJintk) - { -     return(f[k]-f[j]+ (s[k]+c) * (s[k]+c)-(s[j]+c) * (S[J]+C))/(2.0* (s[k]-s[j])); - } - voidDP () - { inL=1; r=0; q[++r]=0; -      for(intI=1; i<=n;i++) to     { +          while(L<r&&slop (q[l],q[l+1]) <=s[i]) l++; -         intt=Q[l]; thef[i]=f[t]+ (s[i]-s[t]-c) * (s[i]-s[t]-C); *          while(L<r&&slop (Q[r],i) <slop (q[r-1],Q[R]) r--; $q[++r]=i;Panax Notoginseng     } - } the intMain () + { AN=read (); L=read (); c=l+1; the      for(intI=1; i<=n;i++) c[i]=read (); +      for(intI=1; i<=n;i++) s[i]=s[i-1]+C[i]; -      for(intI=1; i<=n;i++) s[i]+=i; $ DP (); $printf"%lld\n", F[n]); -     return 0; -}

View Code

"Slope" Toy packing