hdu1024 Max Sum Plus Plus--DP

原题链接： ​​ http://acm.hdu.edu.cn/showproblem.php?pid=1024​​

一：原题内容

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 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 an integer m (m > 0), your task is to find m pairs of i and j which make sum(i ₁, j ₁) + sum(i ₂, j ₂) + sum(i ₃, j ₃) + ... + 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 S ₁, S ₂, S ₃ ... S _n.
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

Hint

二：分析理解

最大m段子序列和问题。

设输入的数组为a[1...n]，从中找出m个段，使者几个段的和为最大

dp[i][j]表示前j个数中取i个段的和的最大值，其中最后一个段包含a[j]。(这很关键)

则状态转移方程为：

dp[i][j]=max{dp[i][j-1]+a[j],max{dp[i-1][t]}+a[j]}    i-1=<t<=j-1

因为dp[i][j]中a[j]可能就自身一个数组成最后一段，或者a[j]与a[j-1]等前面的数组成最后一段。

此题n数据太大，二维数组开不下，而且三重循环，想到状态转移方程后还是困难重重。

想想，二维数组不行的话，肯定要压缩成一维数组：

因为dp[i-1][t]的值只在计算dp[i][j]的时候用到，那么没有必要保存所有的dp[i][j] 1<=i<=m,这样我们可以用一维数组存储。

用pre[j]表示j之前一个状态dp[i-1][]中1---j之间，不一定包含a[j]的最大字段和，然后推dp[i][j]状态时，dp[i][j]=max{pre[j-1]，dp[j-1]}+a[j];

三：AC代码

#include<iostream>  
#include<string.h>
#include<algorithm>  

using namespace std;

int dp[1000010];
int pre[1000010];
int a[1000010];

int main()
{
  int m, n;
  int maxx;

  while (~scanf("%d%d", &m, &n))
  {
    for (int i = 1; i <= n; i++)
      scanf("%d", &a[i]);

    memset(pre, 0, sizeof(pre));

    for (int i = 1; i <= m; i++)
    {
      maxx = -1 << 28;

      for (int j = i; j <= n; j++)
      {
        dp[j] = max(dp[j - 1], pre[j - 1]) + a[j];
        pre[j - 1] = maxx;

        if (maxx < dp[j])
          maxx = dp[j];
      }
    }

    printf("%d\n", maxx);

  }

  return 0;
}