HDU 1024 Max Sum Plus Plus (动态规划 最大M字段和)

 

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

 题目大意：有n的个数，需要你将他划分成m段互不相交的子段，求最大的子段和

　思路：在这里向大家推荐一篇大佬的博客，讲解的清晰明了，让我获益匪浅 https://blog.csdn.net/winter2121/article/details/72848482  Orz

　　大佬的博客写的很清晰，而且附上图解很容易弄懂

　　dp[i][j] 代表前i个数并且以第i个数为末尾划分成了j段的最大和。

　　而dp[i][j] 有两种情况：1. 将第i个数划分到前面这一段取

　　　　　　　　　　　　2.将第i个数单独拿出来作为一段的开始

　　dp[i][j] = max(dp[i-1][j],max(前i个数中的最大和))+a[j];

#include<iostream>
#include<algorithm>
#include<cstring>

using namespace std;
typedef long long LL;
const int INF = 0x3f3f3f3f;
const int maxn = 1000005;
LL n, m;
LL dp[maxn][2], a[maxn];//dp[i][j]代表以第i个数为结尾且分成j段的最大和
int main()
{
    ios::sync_with_stdio(false);
    while (/*cin>>m>>n*/scanf("%lld%lld", &m, &n) != EOF) {
        for (int i = 1; i <= n; i++)scanf("%lld", &a[i]);
            /*cin >> a[i];*/
        for (int i = 0; i <= n; i++)dp[i][0] = dp[i][1] = 0;
        for (int i = 1, k = 1; i <= m; i++, k ^= 1) {
            LL maxpre = -INF; dp[i - 1][k] = -INF;//避免与后面for循环的影响
            for (int j = i; j <= n - m + i; j++) {
                maxpre = max(maxpre, dp[j-1][k ^ 1]);//寻找上一行第i个数之前的最大值
                dp[j][k] = max(dp[j - 1][k], maxpre) + a[j];
            }
        }
        LL ans = -INF;
        for (int i = m; i <= n; i++)
            ans = max(ans, dp[i][m&1]);
        //cout << ans << endl;
        printf("%lld
", ans);
    }
    return 0;
}