HDU 1024 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): 21300    Accepted Submission(s): 7119

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.
 

 

第一次做到滚动数组DP求解，简单来说，考虑到滚动数组的DP是在于如果开成二维数组会过大的长宽，比方说这个题目，如果以dp[i][j]表示前j个数分成i段的话，就有n * m的大小，再有我们发现，每一行的状态都是由上面那一行的状态所决定，不存在跨多行的问题。

对于这个题目，也是不太会做，看了半天题解。首先写出动态转移方程：

dp[i][j]=max(dp[i][j-1]+num[j],dp(i-1,t)+num[j]) 其中-- 未完整表达式 i - 1 <= t <= j - 1

这个比较好理解，就是将j个数分成i份，其中无非分成两种情况，第一种情况是前面与当前的连在一起，第二种情况即两段分开。我们只要找到dp[i][j - 1]和dp[i - 1][t]中的较大的一个加上当前j位置的数，就可以得到dp[i][j]，所以我们需要一个数组pre_max[N]，用pre_max[j - 1]来代表dp[i - 1][t]的最大值。我们将dp[i][t]的值一直存储，就存储在n的位置。然后再每次更新完当前的pre_max以后更新pre_max[n]位置的值。、

具体参考：http://www.cnblogs.com/dongsheng/archive/2013/05/28/3104629.html

代码如下：

/*************************************************************************
	> File Name: Max_Sum_Plus_Plus.cpp
	> Author: Zhanghaoran
	> Mail: chilumanxi@xiyoulinux.org
	> Created Time: Thu 22 Oct 2015 06:14:46 PM CST
 ************************************************************************/

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <cstdlib>

using namespace std;
const int N = 1000010;
int n, m;
int a[N];
int pre_max[N];

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

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

    return pre_max[n];
}

int main(void){
    while(~scanf("%d %d", &m, &n)){
        memset(pre_max, 0, sizeof(pre_max));
        for(int i = 1; i <= n; i ++){
           scanf("%d", &a[i]);
        }
        cout << dp(m, n) << endl;
    }
    return 0;
}