标签：dp

Max Sum Plus Plus Plus

3
2 1 1
1 2 3
4
2 1 2
1 2 3 5
0

5
10

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dp[i][j]表示把前i个数分成j份，得到的和的最大值，则 dp[i][j] = max(dp[i - 1][j], dp[i - seg[j][j - 1] + sum);

#include <map>
#include <set>
#include <list>
#include <queue>
#include <stack>
#include <vector>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

const int N = 1010;
const int M = 22;

int sum[N];
int dp[N][M];
int arr[N], seg[M];
int main()
{
	int n, m;
	while(~scanf("%d", &n), n)
	{
		sum[0] = 0;
		scanf("%d", &m);
		for (int i = 1; i <= m; ++i)
		{
			scanf("%d", &seg[i]);
		}
		for (int i = 1; i <= n; ++i)
		{
			scanf("%d", &arr[i]);
			sum[i] = sum[i - 1] + arr[i];
		}
		memset (dp, 0, sizeof(dp));
		for (int i = 1; i <= n; ++i)
		{
			for (int j = 1; j <= m; ++j)
			{
				dp[i][j] = max(dp[i][j], dp[i - 1][j]);
				if (i > seg[j])
				{
					dp[i][j] = max(dp[i][j], dp[i - seg[j]][j - 1] + sum[i] - sum[i - seg[j]]);
				}
			}
		}
		printf("%d\n", dp[n][m]);
	}
	return 0;
}

hdu1244——Max Sum Plus Plus Plus

标签：dp

原文地址：http://blog.csdn.net/guard_mine/article/details/41846955