uva 10271

DP  状态转移方程 dp[i][j] = min(dp[i-1][j], dp[i-2][j-1] + w))

dp[i][j] 指的是前i个筷子组成j组所花费的最小值 考虑第i个筷子是否参与第j组 （筷子从大到小排序）

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#define MAX 100000000
using namespace std;
int dp[5010][1010];
int a[5010];

int main()
{
    int t;
    scanf("%d", &t);
    while (t--)
    {
        int m, n;
        scanf("%d %d", &m, &n);
        m += 8;
        for(int i = n; i >= 1; i--)
            scanf("%d", &a[i]);
        for(int i = 1; i <= n; i++)
        {
            dp[i][0] = 0;
            for (int j = 1; j <= m; j++)
                dp[i][j] = MAX;
        }
        for(int i = 3; i <= n; i++)
            for(int j = 1; j <= m && i >= j*3; j++)
                dp[i][j] = min(dp[i-1][j], dp[i-2][j-1] + (a[i]-a[i-1]) * (a[i]-a[i-1]));
        printf("%d\n", dp[n][m]);
    }
    return 0;
}