Max Sum Plus Plus

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：在长度为n的数列中选取k个不重叠的区间，使这k区间和之和最大。输出这个最大值。

 

思路：定义dp[i][j]为用前j个数取得i个区间，并且第j个数包含在最后一个区间时的最优解，则有递推式：dp[i][[j]=max(dp[i][j-1]+a[j], dp[i-1][t]+a[j])，即这里有两种情况，

一种是直接在前j-1个数形成i个区间末尾加上a[j]，因为dp[i][j-1]的最后一个区间包含a[j-1]，所以这种情况区间数是不增加的。

第二种情况，是在形成i-1个区间的情况下，以a[j]为一个新区间，这样就有了i个区间，这里i-1<=t<=j-1。

但是如果这么做复杂度会达到O(n^3),需要降低复杂度。

我们发现，dp[i-1][t]最大值可以同样的以dp的思路记录下来，保存在M[]数组内，每次求dp[i][j]时不用在[i-1, j-1]枚举dp[i-1][t]直接拿M[j-1]就行了。

优化一下代码，dp[i-2][j]在后面的计算用不到了，dp[i-1][j]的最大值被存了下来，于是可以把dp[i][j]改成dp[i]节省空间。

 

AC代码：

1 #include
        
          2 #include
         
           3 #include
          
            4 using namespace std; 5 const int MAXN=1e6+10; 6 const long long INF=1e12+10; 7 long long dp[MAXN]; 8 long long a[MAXN],M[MAXN]; 9 int main()10 {11     int k,n;12     while(~scanf("%d %d", &k, &n))13     {14         for(int i=1;i<=n;i++){15             scanf("%lld", &a[i]);16         }17         memset(dp, 0, sizeof(dp));18         memset(M, 0, sizeof(M));19         long long res;20         for(int i=1;i<=k;i++){21             res=-INF;22             for(int j=i;j<=n;j++){23                 dp[j]=max(dp[j-1], M[j-1])+a[j];24                 M[j-1]=res;25                 res=max(res, dp[j]);26             }27         }28         printf("%d\n", res);29             30     }31 }