HDU 1024：Max Sum Plus Plus（DP，最大m子段和）

                                  Max Sum Plus Plus

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 36673    Accepted Submission(s): 13069

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

Sample Output

6

8

Hint

Huge input, scanf and dynamic programming is recommended.

题意

给出一个长度为n的数组，求数组中m个不相交的子段的最大和

---

参考了kuangbin大神的博客还有好多大佬们的博客才算勉强弄懂了。

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

但是需要对这个方程进行优化，具体优化过程看下面各位大佬们的博客吧

传送门

kuangbin

https://www.cnblogs.com/jiangjing/archive/2013/07/25/3214729.html

https://blog.csdn.net/u013187393/article/details/42914165

代码

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <math.h>
#include <limits.h>
#include <map>
#include <stack>
#include <queue>
#include <vector>
#include <set>
#include <string>
#define ll long long
#define ms(a) memset(a,0,sizeof(a))
#define pi acos(-1.0)
#define INF 0x3f3f3f3f
const double E=exp(1);
const int maxn=1e6+10;
using namespace std;
int a[maxn];
int dp[maxn];//dp[j]表示到第j个时候的最大和
int sum[maxn];//记录上个状态的j-1个前的最大值
int main(int argc, char const *argv[])
{
	ios::sync_with_stdio(false);
	int k,n;
	while(cin>>k>>n)
	{
		ms(a);
		ms(dp);
		ms(sum);
		for(int i=1;i<=n;i++)
			cin>>a[i];
		int ans;
		for(int i=1;i<=k;i++)
		{
			ans=INT_MIN;
			for(int j=i;j<=n;j++)
			{
				//当前位置的最大和
				dp[j]=max(dp[j-1]+a[j],sum[j-1]+a[j]);
				//记录上个状态的最大和
				sum[j-1]=ans;
				ans=max(ans,dp[j]);
			}
		}
		cout<<ans<<endl;
	}
	return 0;
}