code1319 Toy Packing

A division of DP, but because of the number of partitions arbitrarily, only one-dimensional array can be

Set Dp[i] Indicates the minimum cost of the first I box (any case)

Dp[i]=min (Dp[u]+cost (u+1,i))

But the complexity of the n<=50000,n side is obviously unacceptable.

Set Choice[i] Array save for each I value, the enumeration of the maximum value of the U-f[i]

Hit the table, found choice[] is monotonous does not descend

Each time for a new I, you only need to start from choice[i-1] when enumerating u

Code:

#include <iostream>#include<cstring>#defineSize 50005using namespacestd;intN;intC[size];Long LongSum[size];Long LongL;Long LongDp[size];intChoice[size];Long LongCostintLintR) {    Long Longx=sum[r]-sum[l-1] + RM; return(x-l) * (X-L);}intMain () {Freopen ("1319.in","R", stdin); Memset (DP,0x7f,sizeof(DP)); CIN>>n>>m;  for(intI=1; i<=n;i++) {cin>>C[i]; Sum[i]=sum[i-1]+C[i]; } dp[0]=0; choice[0]=0;  for(intI=1; i<=n;i++){         for(intu=choice[i-1];u<=i-1; u++){            if(Dp[u]+cost (u+1, i) <=Dp[i]) {Dp[i]=dp[u]+cost (u+1, i); Choice[i]=T; } }} cout<<Dp[n];     Fclose (stdin); return 0;} 

As for better nlgn optimization and n slope optimization, let's say ... sleeping

code1319 Toy Packing