hdu 1024 Max Sum Plus Plus

原题：

>                                     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 … Sx, … Sn (1 ≤ x ≤ n ≤ 1,000,000, -32768 ≤ Sx ≤ 32767). We define a function sum(i, j) = Si + … + Sj (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(im, jm) maximal (ix ≤ iy ≤ jx or ix ≤ jy ≤ jx 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(ix, jx)(1 ≤ x ≤ m) instead. ^_^

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

Output
Output the maximal summation described above in one line.

Sample Input
1 3 1 2 3
2 6 -1 4 -2 3 -2 3

Sample Output
6
8

Hint

Huge input, scanf and dynamic programming is recommended.

原题链接：
http://hdu.hustoj.com/showproblem.php?pid=1024

题意：这题大概的意思是，给出n个数的序列，然后在里面找出m段出来，使得这m段的和最大，每一段可以是一个数字组成，也可以是几个数字组成，但这几个数字必须连续的，上面的Ouput是这样得出来的：
第一个：1 3 1 2 3，m=1，n=3，在这三个数中取一段出来，使其和最大，最大的一段就是1，2，3这段，和为6
第二个：2 6 -1 4 -2 3 -2 3 m=2，n=6，在这六个数中取两段出来，4，-2，3为一段，最后的数字3为一段，这两段的和最大，为8.

java代码：
这里参考了另外一篇博客写的，另外一篇博客是用C语言写的，那里有更详细的图解
http://blog.sina.com.cn/s/blog_677a3eb30100jxqa.html
这篇博客为了节省内存，采用了连滚数组

import java.util.Scanner;public class Main {    private static long[][] dp = new long[2][1000001];    private static long[] a = new long[1000001];    private static int n,m;    private static long maxNum;  //上一行中j之前最大的那个数    private static long res ;   //分成m段最大值，初始化long范围最小值    private static int index;   //两行数组索引下标    public static void main(String[] args) {        Scanner input = new Scanner(System.in);        while (input.hasNext()){            m = input.nextInt();            n = input.nextInt();            index = 1;            res = Long.MIN_VALUE;            /**             * dp[0][]和dp[1][]是相邻的两行，在处理数据中相互转换             */            for(int i=1;i<=n;i++){                dp[0][i] = dp[1][i] = 0;                a[i] = input.nextInt();            }            for(int i=1;i<=m;i++){                dp[index][i] = dp[1 - index][i-1] + a[i];                maxNum = dp[1 - index][i-1];                for (int j=i+1;j<=n-m+i ;j++){                    maxNum = Math.max(maxNum,dp[1-index][j-1]);                    dp[index][j] = Math.max(dp[index][j-1],maxNum) + a[j];                }                index = 1 - index;  //转换两行数据            }            index = 1 - index;            for (int j=m;j<=n;j++){                res = Math.max(dp[index][j],res);            }            System.out.println(res);        }    }}