George and Job

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我是靠谱客的博主外向芹菜，最近开发中收集的这篇文章主要介绍George and Job，觉得挺不错的，现在分享给大家，希望可以做个参考。

概述

The new ITone 6 has been released recently and George got really keen to buy it. Unfortunately, he didn't have enough money, so George was going to work as a programmer. Now he faced the following problem at the work.

Given a sequence of n integers p₁, p₂, ..., p_n. You are to choose k pairs of integers:

[l₁, r₁], [l₂, r₂], ..., [l_k, r_k] (1 ≤ l₁ ≤ r₁ < l₂ ≤ r₂ < ... < l_k ≤ r_k ≤ n; r_i - l_i + 1 = m),

in such a way that the value of sum $\text{[math]}$ is maximal possible. Help George to cope with the task.

Input

The first line contains three integers n, m and k (1 ≤ (m × k) ≤ n ≤ 5000). The second line contains n integers p₁, p₂, ..., p_n(0 ≤ p_i ≤ 10⁹).

Output

Print an integer in a single line — the maximum possible value of sum.

        Sample test(s) 
      
          input 
        
5 2 1
1 2 3 4 5

          output 
        
9

          input 
        
7 1 3
2 10 7 18 5 33 0

          output 
        
61

就是k组数，每组是连续的m个数，求这些数最大和。群里有人推荐这道题，于是来写了一下，题目意思还是很好理解的，很简单的动规思路。连续m个数，那就对每一个位置都算出前m个数的和，然后就是取k次的一个背包。

dp[i][j]=max(dp[i][j-1],dp[i-1][j-m]+d[j]);

i代表取的次数，j代表到第j个数。但是初始化还有一点细节要处理。

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
int a[5005];
long long dp[5005], dp1[5005][5005];
long long max(long long a, long long b)
{
	return a > b ? a : b;
}
int main()
{
	int i, j, m, n, ans, k;
	long long sum = 0, pre;
	cin >> n >> m >> k;
	for (i = 1; i <= n; i++)
		cin >> a[i];
	for (i = 0; i < m; i++)
	{
		dp[i] = 0;
		sum = sum + a[i];
	}
	a[0] = 0; j = 0;
	for (i = m; i <= n; i++)
	{
		sum = sum - a[j];
		sum = sum + a[i];
		dp[i] = sum;
		j++;
	}
	for (i = 0; i < m; i++)
		dp1[1][i] = 0;
	for (i = m; i <= n; i++)
		dp1[1][i] = max(dp1[1][i - 1], dp[i]);
	for (i = 2; i <= k; i++)
	{
		for (j = i * m; j <= n; j++)
		{
			dp1[i][j] = max(dp1[i][j - 1], dp1[i-1][j - m] + dp[j]);
		}
	}
	//for (i = 1; i <= n; i++)
	//	cout << dp[i] << " ";
	//cout << endl;
	//for (i = 1; i <= k; i++)
	//{
	//	for (j = 1; j <= n; j++)
	//		cout << dp1[i][j] << " ";
	//	cout << endl;
	//}
	cout << dp1[k][n] << endl;
	return 0;
}

好久不写动规，这次感觉不错，得写专题了，模拟已经够了。