Hangzhou Electric (Max Sum plus plus)

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Problem Descriptionnow I Think you have got a AC in IGNATIUS.L ' s "Max Sum" problem. To is a brave acmer, we always challenge ourselves to more difficult problems. Now you is faced with a more difficult problem.

Given a consecutive number sequence S 1, S 2, S 3, S 4... S x, ... S N(1≤x≤n≤1,000,000, -32768≤s x≤32767). We define a function sum (i, j) = S I+ ... + S J(1≤i≤j≤n).

Now given a integer m (M > 0), your task is to find m pairs of I and J which make sum (i 1J 1) + SUM (i 2J 2) + SUM (i 3J 3) + ... + sum (i mJ m) Maximal (i x≤i y≤j xOr I x≤j y≤j xis not allowed).

But I ' m lazy, I don't want to the write a Special-judge module, so you don ' t has to output m pairs of I and j, just output th e maximal summation of sum (i xJ x) (1≤x≤m) instead. ^_^

Inputeach test case would begin with a integers m and n, followed by n integers S 1, S 2, S 3... S N.
Process to the end of file.

Outputoutput the maximal summation described above in one line.

Sample Input

1 3 1 2 32 6-1 4-2 3-2 3

Sample Output

68

Ideas:

Classic Dynamic Programming Optimization problem: Set F (i, j) to indicate that the first I number is divided into J, and includes the number of the largest m sub-segment and, then there is the DP equation:
F (i, j) = max {f (i-1, J) + V[i], max {f (k, j-1) + V[i]} (k = j-1 ... i-1)}. You can introduce a secondary array to optimize the transfer. Set g (I, j) to indicate that the number of first I is divided into the maximum sub-segment of J and (note that the number of I may not be in the J-segment), then the recursive relationship is as follows: G (i, j) = Max{g (I-1, J), F (i, j)}, whether or not the number of I is added to transfer such a recursive relationship of F will become: f (i, J) = Max{f (I-1, J), G (I-1, J-1)} + v[i], so the final result is g[n][m], which cleverly optimizes the transfer by introducing an auxiliary array. The implementation can use a one-dimensional array, speed quickly g[i][j] or g[i-1][j] equal, or f[i][j] equal, f[i][j]-a[i] or g[i-1][j-1] equal, or f[i-1][j] equal. Turn into a one-dimensional array to the first row: f[j]=max{f[j],g[j-1]}+a[i]} g[j]=max{g[j],f[j]}.

Code implementation:

Import java.util.*;class main{public    static void Main (string[] args) {        int J;        Scanner sc=new Scanner (system.in);        while (Sc.hasnext ()) {            int m=sc.nextint (); int n=sc.nextint ();            Int[] a=new int[n+1];int[] dp=new int[m+1];int[] dp2=new int[m+1];            for (int i=1;i<=n;i++) {                a[i]=sc.nextint ();            }            DP[1]=A[1];DP 2[1]=a[1];            for (int i=2;i<=n;i++) {                for (j=1;j<=math.min (I, M); j + +) {                    Dp2[j]=math.max (dp2[j]+a[i], dp[j-1]+a[i]) ;                    Dp[j-1]=math.max (Dp[j-1], dp2[j-1]);                }                Dp[j-1]=math.max (Dp[j-1], dp2[j-1]);            }            System.out.println (Dp[m]);}}    

Hangzhou Electric (Max Sum plus plus)