hdoj 1024

Max Sum Plus Plus

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 35117    Accepted Submission(s): 12510

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.

 

题意：输入m，n，然后有n个数，求m段总和最大。

这道题前面想着把它分为m段，然后求这m段的最大子段和。后来越弄越糟，直接放弃这个思路。又想着暴力组合一下各个的和。想不出思路，就去找题解了。（萌新果然还是太嫩了qwq）

 

后面，单纯只看到这个状态转移方程都写不出能a的代码。。。。。

状态转移方程：  dp[i][j]=max(dp[i][j-1]+num[j],dp(i-1,t)+num[j])   其中i-1<=t<=j-1.

#include<cstdio>
using namespace std;
const int maxn=1000006;
int a[maxn],dp[maxn];

int my_max(int x,int y)
{
    return x>y?x:y;
}

int main()
{
    int n,m;
    while( ~scanf("%d%d",&m,&n)){

        for(int i=1;i<=n;i++){
            scanf("%d",&a[i]);
            dp[i]=0;
        }

        int temp;
        for(int i=1;i<=m;i++){
            temp=0;
            for(int j=1;j<=i;j++)
                temp+=a[j];
            dp[n]=temp;

            for(int j=i+1;j<=n;j++){
                temp=my_max( dp[j-1], temp)+a[j];
                dp[j-1]=dp[n];
                dp[n]=my_max( dp[n], temp);
            }
        }

        printf("%d
",dp[n]);
    }
    return 0;
}

https://www.cnblogs.com/dongsheng/archive/2013/05/28/3104629.html

 

#include<cstdio>
#include<climits>
#include<cstring>
#include<iostream>
const int maxn=1000006;
int m,n;
int data[maxn], cur[maxn], pre[maxn];

int Maxsum()
{
    int max_sum;
    for(int i=1;i<=m;i++){
        max_sum=INT_MIN;
        
        for(int j=i;j<=n;j++){
            if(cur[j-1]<pre[j-1])
                cur[j]=pre[j-1]+data[j];
            else 
                cur[j]=cur[j-1]+data[j];

            pre[j-1]=max_sum;
            
            if(max_sum<cur[j])
                max_sum=cur[j];
        }

    }
    return max_sum;
}

int main()
{
    while( ~scanf("%d%d", &m,&n)){
        memset( cur, 0, sizeof cur);
        memset( pre, 0, sizeof pre);

        for(int i=1;i<=n;i++)
            scanf("%d",&data[i]);
        printf("%d
",Maxsum());
    }
    return 0;
}

这个相对好理解。