hdu1024—Max Sum Plus Plus（最大M子段和）

题目链接：传送门

Max Sum Plus Plus

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 34604    Accepted Submission(s): 12363

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 32 6 -1 4 -2 3 -2 3

 

Sample Output

68

Hint

Huge input, scanf and dynamic programming is recommended.

 

解题思路：

我们用dp[i][j]表示前j个数划分为k段的最大和，且以第j个数结尾。

于是有状态转移方程dp[i][j] = max(dp[i][j-1],dp[i-1][k])+data[j]  (i-1 <= k <= j-1)。

这里我们可以用滚动数组对上述方程的空间进行优化。

用r[j]表示dp[i-1][k]的最大值，dp[j]表示dp[i][j]，于是有dp[j] = max(dp[j-1],r[j-1])+data[j]。而r的值可由上一层的dp值得到。

代码：

#include <iostream>
#include <stack>
#include <vector>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>

using namespace std;

typedef long long llt;

const int N = 1000010;
const int M = 510;
const llt INF = 0x3ffffffffffffff;
const int mod = 10000;

llt r[N],dp[N],data[N];

int main()
{
    int m,n; llt c;
    while(~scanf("%d%d",&m,&n)){
        for(int i = 1; i <= n; ++i) scanf("%lld",&data[i]);
        memset(r,0,sizeof(r));
        for(int i = 1; i <= m; ++i){
            c = -INF;
            for(int j = i; j <= n; ++j){
                dp[j] = max(dp[j-1],r[j-1])+data[j];
                r[j-1] = c;
                c = max(c,dp[j]);
            }
        }
        printf("%lld\n",c);
    }
    return 0;
}