Max Sum Plus Plus

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 27299 Accepted Submission(s): 9499

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

Hint

Huge input, scanf and dynamic programming is recommended.

与求最大连续子段略有不同，此题目要求求出m个连续子段的sum，且使得sum尽量大；

考虑状态转移方程：f{i,j}表示前j个元素分为i段的最大值，则有 f{i,j}=max{f{i,j-1},max{i-1,k}}+a[j],(i-1<=k<j);

两种决策：

-1-: 前j个元素分为i段切最后一段以a结尾；

-2-：aj独自占一段，剩下的i-1段选前k（i-1<=k<j)个元素的i-1段中的最大值

n为100w考虑使用滚动数组，pre[j]保存上一个i对应的j的的最大值，dp[j]为当前正在计算的i的j的最大值；

代码：

#include<bits/stdc++.h>
using namespace std;
const int inf=999999999;
int dp[1000005],a[1000005],pre[1000005];
int main()
{
int n,m,i,j,k,t;
while (cin>>m>>n){int maxn=-inf;t=0;
memset(dp,0,sizeof(dp));
memset(pre,0,sizeof(pre));
for (i=1;i<=n;i++) scanf("%d",&a[i]);

for (i=1;i<=m;i++){
maxn=-inf; //每一轮开始都将maxn初始化
for (j=i;j<=n;j++){
dp[j]=max(pre[j-1],dp[j-1])+a[j]; //先使用在更新上一轮的值
pre[j-1]=maxn; //此处很重要，不可写成下文注释的形式，因为一旦那样写下一个j计算时用到的j-1变成了本轮的最大值的意思，概念就变了

if (maxn<dp[j]) maxn=dp[j];
//pre[j]=maxn;
}
//for(int l=0;l<=n;l++) cout<<pre[l]<<" ";
//cout<<endl;
}
cout<<maxn<<endl;
}
return 0;
}