[DP topic-Quadrilateral inequality optimization]51nod 1022

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1021? Stone Merge ? V1 n Heap of stones into a line. It is now necessary to merge the stones into a pile in order. It is stipulated that each of the next 2 stones can be merged into a new pile each time, and a new pile of stones is counted as the cost of the merger. Calculates the minimum cost of merging n heap stones into a heap. ? For example: 1 2 3 4, there are a number of consolidation methods 1 2 3 4 = 3 3 4 (3) = 6 4 (9) = 10 (19) 1 2 3 4 = 1 5 4 (5) = 1 9 (+) + 10 (24) 1 2 3 4 = 1 2 7 (7) = 3 7 (Ten) + 10 (20) ? The total cost in parentheses shows that the first method has the lowest cost and now gives the number of n heap stones, calculating the minimum combined cost. Input Line 1th: N (2 <= n <= 100) Number of 2-n + 1:n heap stones (1 <= a[i] <= 10000) Output Minimum combined cost of output Input example 4 1 2 3 4 Output example 19

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1022? Stone Merge ? V2 N The heap of stones was put into a ring. It is now necessary to merge the stones into a pile in order. it is stipulated that each of the next 2 stones can be merged into a new pile each time, and a new pile of stones is counted as the cost of the merger. Calculates The minimum cost of merging N heap stones into a heap. ? For example: 1 2 3 4 , there are a number of consolidation methods 1 2 3 4 = 3 3 4 (3) = 6 4 (9) = 10 (19) 1 2 3 4 = 1 5 4 (5) = 1 9 (+) + 10 (24) 1 2 3 4 = 1 2 7 (7) = 3 7 (Ten) + 10 (20) ? The total cost in parentheses shows that the first method has the lowest cost and now gives N Count the number of stones, calculate the minimum combined cost. Input Section 1 Line: N ( 2?<=? n?<=?1000) Section 2?-? N?+?1: number of stones in N heap (1?<=? a[i]?<=?10000) Output Minimum combined cost of output Input Example 4 1 2 3 4 Output Example 19

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The first problem is better written, the complexity is N3, but the second problem needs to use the Quadrilateral inequality optimization, the complexity reduced to N2.

"Algorithmic Competition Introduction Classic Training Guide-Rujia" P170 (old version)

In the book, we refer to the optimization method of quadrilateral inequalities when we discuss the optimal ordered binary tree. In general, for a similar state transfer equation:

???? D[I,J] = max{d[i,k?1] + d[k+1,j]} + w[i,j]

The status has O (N2), each decision has O (n), so the time complexity is O (N3).

The following optimizes it to O (N2).

quadrilateral Inequalities (Monge Condition/quadrangle inequality) if for i≤i ' < j≤j ' There is always w[i,j] + w[i ', J ']≤w[i ', j] + W[i, J ' ], called W satisfies the quadrilateral inequality, or w satisfies convexity.

the monotony of the interval containing lattice if for i≤i ' < j≤j ' There is always w[i ', j]≤w[I, J '], called W satisfies monotonicity, i.e. for two intervals q1,q2 (Q2 contains Q1), W[Q1]≤W[Q2].

theorem (F.yao): If W satisfies the quadrilateral inequalities, D also satisfies the quadrilateral inequalities, i.e.

D[I,J] + d[i ', j ']≤d[I ', j] + d[I, J '], I≤i ' ≤j≤j '

Proof slightly.

Further, the convexity of D can introduce the monotonic of decision making . K[I,J] for the decision to take the minimum value for D[i,j

theorem (F.yao):k[i,j]≤k[i,j + 1]≤k[i + 1,j + 1], i≤j, that is, K is incremented in the same column as the peers.

Proof: by symmetry, just prove k[i,j]≤k[i,j + 1]. Kee Dk[i,j] = d[i,k? 1] + d[k,j] +w[i,j], then simply prove to all I < k≤k ' ≤j, if DK '[I,J]≤DK [I,j], then DK ' [I,j + 1] ≤DK [I,j + 1], that is: After the interval is extended a unit, the previous better decision is still good now . In fact, a stronger conclusion can be proved: Dk [i,j]?d k '[i,j]≤d k [i,j +1]?dk ' [i,j +1] (i <k≤k ' ≤j), because when K ' is better in [I,j] , left ≥0, by inequality know right ≥0, so K ' is still better.

For example, set K ' is the optimal value of [i,j], then for its left of any k,k ' on [i,j] better means that K ' is still optimal on [i,j+1], so K ' left of any k on [i,j+1] is still not the maximum value, that is K[i,j + 1]≥k[i,j].

to certify DK [i,j]–dk ' [I,J]≤DK [I,j + 1]–dk ' [I,j + 1] (i < k≤k ' ≤j),

Shift term: DK [i,j] +dk ' [i,j + 1]≤dk [i,j + 1] + Dk ' [i,j].

Expand by definition, both sides eliminate W[I,J] + w[i,j + 1] + d[i,k? 1] + d[i,k 0? 1]:

???????? D[K,J] + d[k ', j + 1]≤d[k,j + 1] + d[k ', j]

This is the convexity of D.

With decision Monotonicity, you can change the program a little, and make decisions from k[i,j?1] to k[i+1,j].

Does this reduce the complexity of the time? When L = j? 1 when fixed,

D[1,l + 1] The decision is k[1,l] ~ k[2,l + 1]

D[2,l + 2] Decision is K[2,L + 1] ~ k[3,l + 2]

D[3,l + 3] Decision is K[3,L + 2] ~ k[4,l + 3]

D[4,l + 4] Decision is K[4,L + 3] ~ k[5,l + 4]

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Combined, when L fixed the total decision was k[1,l] ~ k[n? L+1,n], a total of O (n). Because L has O (n), the total time complexity is reduced to O (n 2).

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Extended

The state transfer equation given above is only a relatively general one, in fact, many state transfer equations satisfy the conditions of optimization of quadrilateral inequalities.

The approximate steps to solve this type of problem are:

2. prove W satisfies the quadrilateral inequalities, here W is a m the number of attachments, shaped like M[i,j]=opt{m[i,k]+m[k,j]+w[i,j]} , most of them must first prove W meet the conditions to further prove m Meet the conditions
4. prove m satisfying quadrilateral inequalities
6. prove S[i,j-1]≤s[i,j]≤s[i+1,j]

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Whether opt is min or Max , regardless of M[i,j]=opt{m[i,k]+m[k,j]+w[i,j]} , each step of the decision is enumerated from k[i,j1] to k[i+1,j]. This will reduce the complexity. As far as 51nod is concerned, so does its steps.

[DP topic-Quadrilateral inequality optimization]51nod 1022