Max Sum Plus Plus

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

Total Submission(s): 22262 Accepted Submission(s): 7484

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

Sample Output

6
8

[题意]:输入一个m，n分别表示成m组，一共有n个数即将n个数分成m组，m组的和加起来得到最大值并输出。

[分析]:

状态dp[i][j]表示前j个数分成i组的最大值。

动态转移方程：dp[i][j]=max(dp[i][j-1]+a[j],max(dp[i-1][k])+a[j]) (0<k<j)

dp[i][j-1]+a[j]表示的是前j-1分成i组，第j个必须放在前一组里面。

max( dp[i-1][k] ) + a[j] )表示的前（0<k<j）分成i-1组，第j个单独分成一组。

但是题目的数据量比较到，时间复杂度为n^3，n<=1000000，显然会超时，继续优化。

max( dp[i-1][k] ) 就是上一组 0....j-1 的最大值。我们可以在每次计算dp[i][j]的时候记录下前j个
的最大值用数组保存下来，这样时间复杂度为 n^2。

[代码]:

 /*

 输入一个m，n分别表示成m组,一共有n个数

 即将n个数分成m组，

 m组的和加起来得到最大值并输出。

 */

 #include <bits/stdc++.h>

 using namespace std;

 const int N=;

 #define INF 0x7fffffff

 int a[N+];

 int dp[N+],Max[N+];

 int main()

 {

     int n,m,maxs;

     while(~scanf("%d%d",&m,&n))

     {

         for(int i=;i<=n;i++)

         {

             scanf("%d",&a[i]);

         }

         memset(dp,,sizeof(dp));

         memset(Max,,sizeof(Max));

         for(int i=;i<=m;i++)

         {

             maxs=-INF;

             for(int j=i;j<=n;j++)

             {

                 dp[j]=max(dp[j-]+a[j], Max[j-]+a[j]);

                 Max[j-]=maxs;

                 maxs=max(maxs, dp[j]);

             }

         }

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

     }

 }

线性DP

巴特西

HDU 1024 Max Sum Plus Plus [动态规划+m子段和的最大值]

Max Sum Plus Plus

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