uva 10271 dp

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 5050;
const int maxk = 1010;
const int INF = 0x3f3f3f3f;
int T, k, n, init[maxn], dp[maxn][maxk];
int main(int argc, char const *argv[])
{
	scanf("%d", &T);
	while (T--)
	{
		scanf("%d%d", &k, &n);
		k += 8;
		for (int i = n; i >= 1; i--)
			scanf("%d", &init[i]);
		memset(dp, INF, sizeof(dp));
		for (int i = 0; i < n; i++)
			dp[i][0] = 0;
		for (int i = 3; i <= n; i++)
			for (int j = 1; j <= k; j++)
				if (i >= j * 3 && dp[i - 2][j - 1] != INF)
					dp[i][j] = min(dp[i - 1][j], dp[i - 2][j - 1] + (init[i] - init[i - 1]) * (init[i] - init[i - 1]));
		printf("%d\n", dp[n][k]);
	}
	return 0;
}

简单的二维dp，有n个数据，给定k，要从中选出k+8个三元组（x，y，z，其中x<=y<=z)，每选一次的代价为（x-y）^2，求最小代价和。

二维，i表示第i个，j表示前j组，选或者不选两种情况，答案保存在dp[N][K];