hdu 1024 Max Sum Plus Plus（最大m字段和）

Max Sum Plus PlusTime Limit:1000MS    Memory Limit:32768KB    64bit IO Format:%I64d & %I64u

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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. 
         

 

题目大意：

已知有n个数，求m段不相交的子段权值之和最大，

状态转移方程：dp[i][j]表示以i为结尾元素的j个子段的数和

dp[i][j]=max(dp[i-1][j]+s[i],max(dp[1][j-1]--dp[i-1][j-1])+s[i]);

其中max(dp[1][j-1]--dp[i-1][j-1])为结尾元素在1到i-1中字段数为j-1的最大值

for(j=1; j<=m; j++)
{
	mmx=-999999999;
	for(i=j; i<=n; i++)
	{
		dp[i]=max(dp[i-1]+s[i],ans[i-1]+s[i]);//其中dp[i-1]表示的是以i-1结尾的元素j个子段的数和， ans[i-1]表示的是前i-1个元素中j-1个子段的数和的最大值。
		ans[i-1]=mmx;放在此处是为了实现ans[i-1]+a[i]中a[i]是一个独立的子段，那么此时就应该用的是j-1段
		if(dp[i]>mmx)
		   mmx=dp[i];
	}
	



#include <iostream>
#include <string.h>
using namespace std;
const int nmax = 1000005;
int dp[nmax], s[nmax], ans[nmax];
int main()
{
	int m, n, i, j;
	while(~scanf("%d%d",&m,&n))
	{
		for(i=1; i<=n; i++)
			scanf("%d", &s[i]);
		memset(dp, 0, sizeof(dp));
		memset(ans, 0, sizeof(ans));
		int mmx;
		for(j=1; j<=m; j++)
		{
			mmx=-999999999;
			for(i=j; i<=n; i++)
			{
				dp[i]=max(dp[i-1]+s[i],ans[i-1]+s[i]);
				ans[i-1]=mmx;
				if(dp[i]>mmx)
					mmx=dp[i];
			}
			
		}
		printf("%d\n",mmx);
	}

}
}