"bzoj1010" [HNOI2008] Toy packing toy

The first time to write the slope optimization of the problem is good ah = =

But the question seems to be a dumb tat,

And then this is the article ... http://wenku.baidu.com/link?url= Uqoesurresum4nve5zachn8kak5hgztj5umfmzegjfqs6uvehq2s8zh7iitt7din47sfxztvy041l5tzdvltl1wsua35umiwfp8gqnpnues

S[i] is the prefix of c[i] and, F[i] represents the minimum cost of a 1th item to the first item.

The easy-to-get DP equation is f[i]=min (f[j]+ (s[i]-s[j]+i-j+1-l) ^2).

Set T[i]=s[i]+i, then F[i]=min (f[j]+ (t[i]-t[j]+1-l) ^2.

Set M=T[I]-L-1, then F[i]=min (f[j]+ (m-t[j]) ^2).

When I make a decision, J,k is set to 2 general decision points. Set j<k.

If K is superior to J, there is

f[k]+ (M-t[k]) ^2<f[j]+ (M-t[j]) ^2

Unfold, get

F[k]+m^2-2*m*t[k]+t[k]^2<f[j]+m^2-2*m*t[j]+t[j]^2

Simplify, get

((f[k]+t[k]^2)-(f[j]+t[j]^2))/(T[k]-t[j]) <2*m=2* (S[I]+I-L-1)

Make slope (i,j) = (F[I]+SQR ((double) s[i]+i)-f[j]-sqr ((Double) s[j]+j)/(s[i]+i-s[j]-j)

2 important conclusions of slope optimization (for this problem):

1) j<k, if Slope (j,k) <2* (S[I]+I-L-1), then K ratio J Excellent. (PS: The left side of the inequality is only a j,k-related equation, and the right side of the inequality is only the formula related to i)

2) J<k<l, if Slope (j,k) >slope (k,l), then K can be shed.

Monotonic Queue optimization DP pseudo-code:

 for (int i=1; i<=n;++i) {    while (H +1<t && slope (q[h],q[h+ 1]) <, ...) h++;    F[i]=...;      while (H +1<t && slope (q[t-2],q[t-1]) >slope (q[t-1],i)) t-- ;    Q[t++]=i;}

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"bzoj1010" [HNOI2008] Toy packing toy