Hdu 2059 Tortoise and rabbit race (DP)

Hdu 2059 Tortoise and rabbit race (DP)

 

It is not clear that the remaining power of electric vehicles cannot be used, and each time it is full.

 

Question Analysis:

For dynamic planning problems, you can perform DP on the site, plus a total of n + 2 sites at the start and end points. The start point indicates that the site is 0, and dp [n + 1] indicates the optimal time of the tortoise, that is, the shortest time, the dynamic conversion equation is: dp [I] = min (dp [I], dp [j] + t1); t1: the optimal (minimum) time from j to I.

 

AC code:

 

/*** @ Xiaoran * DP. The question is unclear and the remaining energy cannot be used * dp [I] = min (dp [I], dp [j] + cost ), cost: time from j --> I */# include
 
  
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             # Define LL long longusing namespace std; const double DOB_MAX = 1000000.0; int path [110]; double dp [110]; int main () {int l, n, c, t, vr, vt1, vt2; while (scanf (% d, & l) = 1) {scanf (% d, & n, & c, & t ); scanf (% d, & vr, & vt1, & vt2); for (int I = 1; I <= n; I ++) {scanf (% d, & path [I]);} path [n + 1] = l; dp [0] = 0; double t1, t2; for (int I = 1; I <= n + 1; I ++) {dp [I] = DOB_MAX; for (int j = 0; j
            
              = Len) t1 = len * 1.0/vt1; // else t1 = c * 1.0/vt1 + (len-c) * 1.0/vt2; // use an electric car to drive c, and the remaining pedal if (j) t1 + = t; // calculate the fuel time. // 2. Do not refuel t2 = len * 1.0/vt2; t1 = min (t1, t2); // minimum cost from I to j dp [I] = min (dp [I], dp [j] + t1 );}} if (dp [n + 1]