HDU1024——Max Sum Plus Plus

Max Sum Plus Plus

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7919    Accepted Submission(s): 2624

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.

 

 

Author

JGShining（极光炫影）

 

 

题意：给定一个数组，求其分成m个不相交子段和最大值的问题。

状态方程：方程 dp[i][j]=Max(dp[i][j-1]+a[j] , max( dp[i-1][k] ) + a[j] ) 0<k<j。

代码：

#include <stdio.h>
#include <string.h>
#define  N 1000010
#define inf 9999999

int dp[N], pre[N], num[N], max;

int Max(int a, int b)
{
    return a > b ? a : b;
}
int main()
{
    int i, j;
    int n, m;
    while(scanf("%d%d", &m, &n) !=EOF)
    {
        for(i=1;i<=n;i++)
        {
            scanf("%d", num+i);
        }
        memset(dp, 0, sizeof(dp));
        memset(pre, 0, sizeof(pre));
        for(i=1;i<=m;i++)
        {
            max = -inf;
            for(j=i;j<=n;j++)
            {
                dp[j] = Max(dp[j-1]+num[j], pre[j-1]+num[j]);
                pre[j-1] = max;
                if(max < dp[j])
                {
                    max = dp[j];
                }
            }
        }
        printf("%d\n", max);
    }
    return 0;
}