HDU-1024 Max Sum Plus Plus(dp)

Max Sum Plus Plus

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 26025    Accepted Submission(s): 9055

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 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.

题意：n个数，求m段连续序列的最大和。

思路：

dp[i][j]表示前j项中i个子段和的最大值
考虑第j个数，是包含在第i组里面，还是自己独立成组作为第一个

所以dp[i][j] = max(dp[i][j-1],dp[i-1][k]);

然后滚动数组优化空间，用i-1，j的最大值作为

/*
dp[i][j]表示前j项中i个子段和的最大值
考虑第j个数，是包含在第i组里面，还是自己独立成组作为第一个

*/
#include
#include
#include
#include
using namespace std;

const int INF = 0x3f3f3f3f;
const int N = 1000005;
int s[N],dp[N],maxn[N];
int main()
{
    int m,n;
    while(~scanf("%d%d",&m,&n))
    {
        for(int i = 1;i <= n;i++)
            scanf("%d",&s[i]);

        memset(dp,-0x3f,sizeof(dp));
        memset(maxn,0,sizeof(maxn));
        int ans;
        for(int i = 1;i <= m;i++)
        {
            ans = -INF;
            for(int j = i;j <= n;j++)
            {
                dp[j] = max(dp[j-1],maxn[j-1]) + s[j];
                maxn[j-1] = ans;
                ans = max(ans,dp[j]);
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}