hdu 1024 Max Sum Plus Plus

Max Sum Plus Plus

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Problem Description

Now I think you have got an AC in Ignatius.L's "Max Sum" problem. To be a brave ACMer, we always challenge ourselves to more difficult problems. Now you are faced with a more difficult problem.

Given a consecutive number sequence S₁, S₂, S₃, S₄ ... 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 an integer m (m > 0), your task is to find m pairs of i and j which make sum(i₁, j₁) + sum(i₂, j₂) + sum(i₃, j₃) + ... + sum(i_m, j_m) maximal (i_x ≤ i_y ≤ j_x or i_x ≤ j_y ≤ j_x is not allowed).

But I`m lazy, I don't want to write a special-judge module, so you don't have to output m pairs of i and j, just output the maximal summation of sum(i_x, j_x)(1 ≤ x ≤ m) instead. ^_^

 

Input

Each test case will begin with two integers m and n, followed by n integers S₁, S₂, S₃ ... S_n.
Process to the end of file.

 

Output

Output 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
Hint
Huge input, scanf and dynamic programming is recommended.
 

 

Author

JGShining（极光炫影）

 

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题解：

状态dp[i][j]

有前j个数，在选用第j个数的前提下，组成i组的和的最大值。

决策： 第j个数，是在第包含在第i组里面，还是自己独立成组。

方程 dp[i][j]=Max(dp[i][j-1]+a[j] , max( dp[i-1][k] ) + a[j] ) 0<k<j

Code：

import java.math.BigInteger;import java.util.Arrays;import java.util.Scanner;public class Main {public static int maxn = 1000000 + 20;public static void main(String args[]) {Scanner Cin = new Scanner(System.in);int[] num = new int[maxn];int[] dp = new int[maxn];int[] maxs = new int[maxn];int maxx = -0x7fffffff,n,m;while (Cin.hasNext()) {m = Cin.nextInt();n = Cin.nextInt();Arrays.fill(dp, 0);Arrays.fill(maxs, 0);for (int i = 1; i <= n; i++) {num[i] = Cin.nextInt();}for (int i = 1; i <= m; i++) {//i表示分成i组maxx = -0x7fffffff;for (int j = i; j <= n; j++) {//j表示前j个数，合起来就是前j个数(取j的前提下)分成i组最大值是dp[j];dp[j] = Math.max(dp[j - 1] + num[j], maxs[j - 1] + num[j]);//maxs[j - 1]的值是dp[x](1 <= x <= j - 1)的最大值,也就是前x个元素(x从1到j - 1)分成i - 1组的最大值;maxs[j - 1] = maxx;//当前j - 1分成i组的最大值现在才更新，所以上一行是分成i - 1组 的最大值maxx = Math.max(maxx, dp[j]);//作为缓存，存的是1~j中分成i组的最大值}}System.out.println(maxx);}}}