HDU2059龟兔赛跑

　　原题链接：http://acm.hdu.edu.cn/showproblem.php?pid=2059

　　典型的动态规划，这个题没有想太复杂，直接用最简单的dp过了，分类说是数塔变种，但是没有找到思路。所以还是朴素的dp过了，wa了三次，后面还是ac了。hdu的数据确实没有poj的强。说下我的思路，代码写的很挫，肯定没人能看懂。分为两种情况，1、一次油也不加。2、至少加一次油，设d[i][j]为到第i个加油站加了j次油的最快时间。1<=i<=n, 1<=j<=i;代码如下，可供对拍：

#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <set>
#include <string.h>

using namespace std;

#define inf 0x3f3f3f3f
int n;
double l, c, t, vr, vt1, vt2;
double d[105], a[105][105];

bool cal(){
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= n; j++)
            a[i][j] = inf;
    for (int i = 1; i <= n; i++){
        if (d[i] <= c)
            a[i][1] = d[i] / vt1 + t;
        else a[i][1] = c / vt1 + (d[i] - c) / vt2 + t;
    }

    for (int i = 1; i <= n; i++){
        for (int j = 2; j <= i; j++){

            double maxf = 1000000000.0;

            for (int k = 1; k < i; k++){

                double time = a[k][j - 1];
                double dis = d[i] - d[k];

                if (dis <= c)time += dis / vt1;
                else time += (dis - c) / vt2 + c / vt1;

                if (time < maxf)maxf = time;
            }
            a[i][j] = maxf + t;
        }
    }
    double minf = 1000000000.0;

    for (int i = 1; i <= n; i++){

        double time = 100000000.0;
        for (int j = 1; j <= i; j++){
            if (time > a[i][j])time = a[i][j];
        }

        double dis = l - d[i];
        if (dis <= c)time += dis / vt1;
        else time += c / vt1 + (dis - c) / vt2;

        if (minf > time)minf = time;
    }

    double temp;
    if (l <= c)temp = l / vt1;
    else temp = c / vt1 + (l - c) / vt2;

    minf = min(minf, temp);

    if (minf > l / vr)return true;
    return false;
}

int main()
{
    while (cin>>l){
        cin>>n>>c>>t;
        cin>>vr>>vt1>>vt2;
        d[0] = 0;
        for (int i = 1; i <= n; i++){
            cin>>d[i];
        }
        if (cal())
            cout<<"Good job,rabbit!"<<endl;
        else cout<<"What a pity rabbit!"<<endl;
    }
	return 0;
}