Codeforces Round #267 (Div. 2) C. George and Job

最新推荐文章于 2020-08-12 15:14:57 发布

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本文介绍了一道Codeforces上的动态规划题目，旨在通过选取多个固定长度的区间来获得最大和。文章给出了清晰的解题思路及代码实现。

动态规划真是博大精深，虽然这题绝对是dp中的水题，但是在比赛的时候真的没有想到解法。

嗯，积累。

题目链接：http://codeforces.com/contest/467/problem/C

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, ..., pn. You are to choose k pairs of integers:

[l1, r1], [l2, r2], ..., [lk, rk] (1 ≤ l1 ≤ r1 < l2 ≤ r2 < ... < lk ≤ rk ≤ n; ri - li + 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, ..., pn (0 ≤ pi ≤ 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

题意：有n个数字组成的一个序列，从该序列中选取k个区间，每个区间的长度都确定为m，其中k*m<=n，求这k个长度为m的区间的数字之和最大的值。

思路：动态规划，令dp[i][j]表示前i个数选取j对（构成区间[r1,l1]则r1和l1为一对）能够得到的最大值，对于第i个数，有选取和不选取之分，状态转移方程：dp[i][j] = max{dp[i-1][j], dp[i-m][j-1]+sum[i]-sum[i-m]}, 其中sum[i]表示前i个数之和

代码：

#include <iostream>
#include <string>
#include <cstring>
#include <cmath>
#define MAX 5010
using namespace std;

__int64 dp[MAX][MAX];

__int64 max(__int64 a, __int64 b)
{
	return a > b ? a : b;
}

int main()
{
	int n, m, k;
	cin >> n >> m >> k;
	int p[MAX];
	for (int i = 1; i <= n; i++)
		cin >> p[i];

	__int64 sum[MAX];
	memset(sum, 0, sizeof(sum));
	for (int i = 1; i <= n; i++)
		sum[i] = sum[i-1]+p[i];

	for (int i = m; i <= n; i++)
		for (int j = 1; j*m<=i && j<=k; j++)
			dp[i][j] = max(dp[i-1][j], dp[i-m][j-1]+sum[i]-sum[i-m]);
	cout << dp[n][k] << endl;
	//system("pause");
	return 0;
}