Four-Channel simple DP

DP class Topic Find sub-problem (state), and then find the transfer equation, OK

#DP#likes Matrixchain#According to both Point ' s distance to recurrenceclassSolution:#@return A string    defLongestpalindrome (self, s): Length=Len (s) P= [[0 forColinchRange (length)] forRowinchrange (length)] forIinchRange (len (s)): P[i][i]= 1if(i< (LENGTH-1) andS[i] = = s[i+1]): Maxlenth= 2P[i][i+1] = 1Start=I forIinchRange (3,length+1):             forJinchRange (length-i+1): K= j + i-1if(S[j]==s[k] andP[J+1][K-1] = = 1): Maxlenth=I p[j][k]= 1Start=Jreturns[start:start+Maxlenth]if __name__=='__main__': S=solution () test_str='ABCBA'    PrintS.longestpalindrome (TEST_STR)

#dp least number coins#one:overlapping Subproblem#two:optimal substructureclassSolution:#@return A string    defLeast_number_coin (self, S, v):#First initializationMin_number = [1000000]* (s+1) min_number[0]=0#and then,recurrence        #Time need O (n^2), and additional time O (n) to save the Subproblem ' s temp solution         forIinchRange (1,s+1):             forJinchRange (len (v)):PrintI,v[j]if(v[j]<=i and(Min_number[i-v[j]]+1 <Min_number[i]): Min_number[i]= Min_number[i-v[j]]+1PrintMin_numberreturnMin_number[s]if __name__=='__main__': S=solution () money= 11v= [1,3,5]    PrintS.least_number_coin (MONEY,V)

#DP#Time O (n^2), addtional O (n) to save temp resultclassSolution:#@return A string    defLIS (self,v):Printv D= [1]*(Len (v))PrintD forIinchRange (len (v)): forJinchRange (i):Printi,jif(V[j]<=v[i] andD[j]+1>D[i]): D[i]= D[j]+1PrintDPrintMax (d)if __name__=='__main__': S=solution () v= [5,3,4,8,6,7] S.lis (v)

#DPclassSolution:#@return A string    defmax_number_apples (self, apples, N, m):PrintApples s= [[0 forColinchRange (m)] forRowinchrange (n)] forIinchrange (n): forJinchRange (m):if(i==0 andj==0): S[i][j]=Apples[0][0]elif(i==0 andJ>0): S[i][j]= s[i][j-1]+Apples[i][j]elif(j==0 andI>0): S[i][j]= S[i-1][j] +Apples[i][j]Else:                    if(s[i-1][j]>s[i][j-1]): S[i][j]= S[i-1][j] +Apples[i][j]Else: S[i][j]= S[i][j-1] +Apples[i][j]PrintsreturnS[n-1][m-1]if __name__=='__main__': S=solution () n= 3m= 4Apples= [[0 forColinchRange (m)] forRowinchrange (n)] K= 1 forIinchrange (n): forJinchRange (m): Apples[i][j]=k K= k + 1s.max_number_apples (apples,n,m)

One:initialization (Initialize)

Two:recurrence (Recursive)

Polynomial time, additional storage space required

Four-Channel simple DP