hdu_2059 龟兔赛跑

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

dp[i]表示乌龟到达第i个充电站时最少花费时间
到第 i 个充电站后，从起点开始遍历到第 i-1 个充电站，得到最少花费时间

dp[i] = min(dp[j] + time(j-->i) ) + t

                        ì  len / vt1,   len <= c
time(j-->i) =     í
                        î c / vt1 + (len-c) / vt2,   len > c

(len = p[i] - p[j])

01 #include <cstdio>
02 #include <cstring>
03 using namespace std;
04
05 const int N = 110;
06 int p [N ];
07 double dp [N ];
08
09 int main()
10 {
11     int L , n , c , t , vr , vt1 , vt2;
12     while ( scanf( "%d" , & L) != EOF)
13     {
14         memset( dp , 0 , sizeof( dp));
15         scanf( "%d%d%d%d%d%d" , &n , & c , & t , & vr , & vt1 , & vt2);
16         for ( int i = 1; i <= n; i ++)
17             scanf( "%d" , &p [ i ]);
18         p [n + 1 ] = L;
19         for ( int i = 1; i <= n + 1; i ++)
20         {
21             double min = 99999999999.0;
22             for ( int j = 0; j < i; j ++)
23             {
24                 int len = p [ i ] - p [ j ];
25                 double time = dp [ j ] + ( len > c ? ( c * 1.0 / vt1 + ( len - c) * 1.0 / vt2) : ( len * 1.0 / vt1));
26                 if ( j)     //起点不用等待充电
27                     time += t;
28                 min = time < min ? time : min;
29             }
30             dp [ i ] = min;
31         }
32         if ( dp [n + 1 ] < L * 1.0 / vr)
33             printf( "What a pity rabbit! \n ");
34         else
35             printf( "Good job,rabbit! \n ");
36     }
37 }