这里面，time[i]表示i船回到港口开始继续运送car的最短时间。
同时设i次最后一次运送了j个car，这样自问题则为time[i-j]了。

#include <iostream>
#include <fstream>
#define in cin
using namespace std;

const int MAX = 1441;
const int INFINITE = 100000000;

int arr[MAX], time[MAX], n, t, m, solution[MAX];

int gettime ( int num )
{
	if ( num <= 0 )
		return 0;

	if ( time[num] > 0 )
		return time[num];

	if ( num == 1 )
	{
		solution[num] = 1;
		time[num] = arr[num] + 2 * t;
		return time[num];
	}

	int min = INFINITE;
	for ( int i = 1; i <= n; i++ )
	{
		int v = gettime ( num - i );

		time[num - i] = v;

		if ( v <= arr[num] )
			v = arr[num] + 2 * t;
		else
			v += 2 * t;

		if ( min > v )
		{
			solution[num] = i;
			min = v;
		}
	}

	time[num] = min;

	return time[num];
}

int main ()
{
	int c, i;
    ifstream in ( "2336.txt" );

	for ( in >> c; c > 0; c-- )
	{
		in >> n >> t >> m;

		for ( i = 1; i <= m; i++ )
			in >> arr[i];

		memset ( solution, 0, sizeof ( solution ) );
		memset ( time, 0, sizeof ( time ) );

		int cost = gettime ( m ), count = 0;
		while ( m > 0 )
		{//cout<<solution[m] << " ";
			count++;
			m -= solution[m];
		}//cout<<endl;
		cout << cost - t << " " << count << endl;
	}

	return 0;
}