HDU1024 DP的优化 最大M子段和问题

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.

不加优化：

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

然后发现k的范围【i-1，j-1】之间可以直接记录一个Maxp

emmmmm，以前做过还是搞忘了

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

         for(i=;i<=m;i++) {
           Maxp=dp[(i-)%][i-];
           dp[i%][i]=dp[(i-)%][i-]+a[i];
           for(j=i+;j<=n-m+i;j++){
                if(dp[(i-)%][j-]>Maxp) Maxp=dp[(i-)%][j-];
                dp[i%][j]=dp[i%][j-]+a[j];
                if(Maxp+a[j]>dp[i%][j]) dp[i%][j]=Maxp+a[j];
         }
        }
         for(i=m;i<=n;i++)
           if(dp[m%][i]>Max) Max=dp[m%][i];
         printf("%d\n",Max);
    }
    return ;
}

至于此题的数据范围，呵呵，不存在的。