hdu2059 龟兔赛跑

来源：互联网发布：triz矛盾矩阵表编辑：程序博客网时间：2024/05/01 18:53

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;}