hdu2059 龟兔赛跑

解题思路：动态规划

状态转移方程：

d[i] = max(d[k] + TT) (0<j<i)

TT : k点充电，从k到i点距离时间

总共n+2点，n个充电站+起点+终点

AC代码：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <cmath>
#include <algorithm>
using namespace std;

#define clr(p,v) memset(p,v,sizeof(p))
const int maxn = 110 ;

int n, m, C, T;
int len, vr, vt, vt2;
int a[maxn];
double d[maxn];

int main()
{
    while (~scanf("%d", &len))
    {
        //Input
        scanf("%d%d%d", &n, &C, &T);
        scanf("%d%d%d", &vr, &vt, &vt2);
        a[0] = 0;
        for (int i=1; i<=n; ++i) scanf("%d", &a[i]);
        a[++n] = len;
        //Caculate
        double tVR = 1.0*len/vr;
        d[0] = 0.0;
        for (int i=1; i<=n; ++i)
        {
            int mi = min(a[i], C);
            d[i] = 1.0*mi/vt + 1.0*(a[i]-mi)/vt2;
            for (int j=i-1; j>0; --j)
            {
                int tt = a[i] - a[j];
                int mi = min(tt, C);
                double tLeft = 1.0*mi/vt+1.0*(tt-mi)/vt2;
                d[i] = min(d[i], d[j]+T+tLeft);
            }
        }
        //output
        if (tVR < d[n]) puts("Good job,rabbit!");
        else puts("What a pity rabbit!");
    }
    return 0;
}