sicily 1264. Atomic Car Race 动态规划

//这题做了一晚
//很简单的dp水题，一开始超时，优化后又WA，结果发现是输出精度不够，至少要输出小数点后4位
//先要预处理一下，不然dp过程中会出现很多重复计算
//设dp[i]为完成路程a[1]+...+a[i]的最小时间，cost[i][j]为从站i换轮胎后，到站j所需的时间
//dp[i] = min(dp[k]+cost[k][i]+b), 0 < k < i
//dp[0] = 0;
#include <iostream>
#include <cstdio>
using namespace std;

const int N = 105;
const int M = 10005;

const int INF = 1000000000;

double dp[N];
double cost[N][N];
double total[M];

int record[N];
int a[N];
int n, r;
double v,e,f,b;

double cal(int x)
{
	if (x >= r)
		return 1/(v- e*(x-r));
	else
		return 1/(v - f*(r-x));
}

int main()
{
	a[0] = 0;
	int maxx, mini;

	while (cin >> n && n)
	{
		maxx = -INF;
		for (int i = 1; i <= n; i++)
		{
			cin >> a[i];
			dp[i] = INF;
			maxx = max(maxx, a[i]);
		}
		cin >> b;
		cin >> r >> v >> e >> f;

		total[0] = cal(0);
		for (int i = 1; i <= maxx; i++)
			total[i] = total[i-1] + cal(i);	
		
		for (int i = 0; i <= n; i++)
			for (int j = i+1; j <= n; j++)
				cost[i][j] = total[a[j]-a[i]-1];		
			
		dp[0] = 0;
		record[0] = 0;
		for (int i = 1; i <= n; i++)
			for (int k = 0; k < i; k++)		
				if (k != 0)
					dp[i] = min(dp[i], dp[k]+cost[k][i]+b);
				else
					dp[i] = min(dp[i], dp[k]+cost[k][i]);		
		
		printf("%.4lf\n", dp[n]);		
	}
	return 0;
}