hdoj 1024

Max Sum Plus Plus

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 35117    Accepted Submission(s): 12510

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

6

8

Hint

Huge input, scanf and dynamic programming is recommended.

 

题意：输入m，n，然后有n个数，求m段总和最大。

这道题前面想着把它分为m段，然后求这m段的最大子段和。后来越弄越糟，直接放弃这个思路。又想着暴力组合一下各个的和。想不出思路，就去找题解了。（萌新果然还是太嫩了qwq）

 

后面，单纯只看到这个状态转移方程都写不出能a的代码。。。。。

状态转移方程：  dp[i][j]=max(dp[i][j-1]+num[j],dp(i-1,t)+num[j])   其中i-1<=t<=j-1.

 1 #include<cstdio>
 2 using namespace std;
 3 const int maxn=1000006;
 4 int a[maxn],dp[maxn];
 5 
 6 int my_max(int x,int y)
 7 {
 8     return x>y?x:y;
 9 }
10 
11 int main()
12 {
13     int n,m;
14     while( ~scanf("%d%d",&m,&n)){
15 
16         for(int i=1;i<=n;i++){
17             scanf("%d",&a[i]);
18             dp[i]=0;
19         }
20 
21         int temp;
22         for(int i=1;i<=m;i++){
23             temp=0;
24             for(int j=1;j<=i;j++)
25                 temp+=a[j];
26             dp[n]=temp;
27 
28             for(int j=i+1;j<=n;j++){
29                 temp=my_max( dp[j-1], temp)+a[j];
30                 dp[j-1]=dp[n];
31                 dp[n]=my_max( dp[n], temp);
32             }
33         }
34 
35         printf("%d\n",dp[n]);
36     }
37     return 0;
38 }

https://www.cnblogs.com/dongsheng/archive/2013/05/28/3104629.html