HDU1024 DP的优化最大M子段和问题

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本文探讨了MaxSumPlusPlus问题，这是一种比经典MaxSum更复杂的动态规划问题。任务是在给定的数列中找到m对下标，使得这些下标的子序列和之总和最大。文章提供了两种实现思路及代码示例，帮助读者理解如何高效解决此类问题。

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Max Sum Plus Plus

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 31583 Accepted Submission(s): 11174

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.

只能说明自己还是太渣。。。

不加优化：

#include<cstdio>
#include<cstdlib>
#include<iostream>
#include<memory.h>
using namespace std;
int dp[2][1000010],a[1000010];
int main()
{
    int n,m,j,i,k,Max;
    while(~scanf("%d%d",&m,&n)){
        Max=0;
        memset(dp,0,sizeof(dp));
        for(i=1;i<=n;i++) scanf("%d",&a[i]);
        for(i=1;i<=m;i++) 
         for(j=m;j<=n;j++){
                dp[i%2][j]=dp[i%2][j-1]+a[j];
                for(k=i-1;k<=j-1;k++)
                 if(dp[(i-1)%2][k]+a[j]>dp[i%2][j]) dp[i%2][j]=dp[(i-1)%2][k]+a[j];
                if(i==m&&dp[i%2][j]>Max) Max=dp[i%2][j];
         }
         printf("%d\n",Max);
    }
    return 0;
}

然后发现k的范围【i-1，j-1】之间可以直接记录一个Maxp

emmmmm，一起做过还是搞忘了

#include<cstdio>
#include<cstdlib>
#include<iostream>
#include<memory.h>
using namespace std;
int dp[2][1000010],a[1000010];
int main()
{
	int n,m,j,i,k,Max,Maxp;
	while(~scanf("%d%d",&m,&n)){
    	Max=-1000000001;
		for(i=1;i<=n;i++) scanf("%d",&a[i]);
		for(i=1;i<=n;i++) dp[0][i]=dp[1][i]=0;
		
		 for(i=1;i<=m;i++) {
		   Maxp=dp[(i-1)%2][i-1];
		   dp[i%2][i]=dp[(i-1)%2][i-1]+a[i];
		   for(j=i+1;j<=n-m+i;j++){
				if(dp[(i-1)%2][j-1]>Maxp) Maxp=dp[(i-1)%2][j-1];
				dp[i%2][j]=dp[i%2][j-1]+a[j];
			    if(Maxp+a[j]>dp[i%2][j]) dp[i%2][j]=Maxp+a[j];
		 }
		}
		 for(i=m;i<=n;i++)
		   if(dp[m%2][i]>Max) Max=dp[m%2][i];
		 printf("%d\n",Max);
	}
	return 0;
}

至于此题的数据范围，呵呵，不存在的。