Max Sum Plus Plus
Time Limit: 2000/1000 MS
(Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5360 Accepted Submission(s): 1773
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
最大和子序列的进化版 , 要求从一序列中取出若干段 , 使得这几段的和最大 .
设 dp[i][j] 为前 j 个数字分成 i 段的最大和 .
转移方程为 :
dp[i][j]=max(dp[i][j-1],max(dp[i-1][k]))+a[j](i-1<=k<=j-1)
其表达的意义就两个不同的决策 : 前者表示与 j-1 所在的一段合并成一段 , 后者表示以
a[j] 为首开始第 i 段 .
下面的代码实现的时候用了滚动数组节约空间 , 可以参考一下 .
代码如下 :