2014-2015 ACM-ICPC, NEERC, Moscow Subregional Contest F. Friends

F. Friends
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

Three friends Alex, Dmitry and Petr want to go abroad for a better life. They start at 09:00 in the morning in their home city A and want to be in the foreign city B as soon as possible.

Alex wants to start immediately and decided to use a car and a highway. He has to drive directly to the customs point C, then spends Dminutes there to pass the customs, and then drives directly to the city B. He drives the car with a speed of V kilometers per hour.

Dmitry has bought an airplane ticket. His plane departs at time T from the city A and flies for some time F directly to the city B.

Petr felt adventurous, so he decided to take his neighbor's tractor and drive straight to the city B with a speed of W kilometers per hour. If Petr's path lies through the customs point C, he spends D minutes there to pass the customs too. Otherwise, he does not stop during the whole trip.

Now they wonder who will be the first in the city B and can cook a welcoming dinner for all of them. We assume that friends live on a plane, and the distance between points is the common Euclidean distance.

Input

The first line of input contains six integers XA, YA, XB, YB, XC and YC, the coordinates in kilometers of cities A and B and the customs point C respectively (0 ≤ XA, YA, XB, YB, XC, YC ≤ 1000). It is guaranteed that the customs point C is closer to A than the city B and closer to B than the city A.

The second line contains two integers D and V, the time to pass the customs in minutes and the speed of Alex's car in kilometers per hour respectively (0 ≤ D ≤ 1000, 40 ≤ V ≤ 100).

The third line contains the departure time T and the duration F of Dmitry's flight, both in the format "HH:MM". It is guaranteed that the departure time is correct, later than 09:00 and earlier than 24:00. The flight duration is greater than 00:00 and less than 24:00.

The fourth line contains one integer W, the speed of Petr's tractor in kilometers per hour (10 ≤ W ≤ 50).

Output

Output the name of the first friend in the city B. It is guaranteed that the absolute difference between arrival times for each pair of friends will be at least one second.

Sample test(s)
input
0 0 10 0 5 5
60 100
10:00 00:05
10
output
Petr
input
0 0 100 0 50 50
0 100
09:01 02:10
50
output
Alex
input
0 0 1000 0 500 500
0 100
10:00 02:10
33
output
Dmitry

 题意：看题目吧

分析：就是暴力算出所有答案就可以了。

非常简单的暴力题。

 

  1 #include <cstdio>
  2 #include <cstring>
  3 #include <cstdlib>
  4 #include <cmath>
  5 #include <ctime>
  6 #include <iostream>
  7 #include <algorithm>
  8 #include <map>
  9 #include <set>
 10 #include <vector>
 11 #include <deque>
 12 #include <queue>
 13 using namespace std;
 14 typedef long long LL;
 15 typedef double DB;
 16 #define Rep(i, n) for(int i = (0); i < (n); i++)
 17 #define Repn(i, n) for(int i = (n)-1; i >= 0; i--)
 18 #define For(i, s, t) for(int i = (s); i <= (t); i++)
 19 #define Ford(i, t, s) for(int i = (t); i >= (s); i--)
 20 #define rep(i, s, t) for(int i = (s); i < (t); i++)
 21 #define repn(i, s, t) for(int i = (s)-1; i >= (t); i--)
 22 #define MIT (2147483647)
 23 #define MLL (1000000000000000000LL)
 24 #define INF (1000000001)
 25 #define mk make_pair
 26 #define ft first
 27 #define sd second
 28 #define clr(x, y) (memset(x, y, sizeof(x)))
 29 #define sqr(x) ((x)*(x))
 30 #define sz(x) ((int) (x).size())
 31 #define puf push_front
 32 #define pub push_back
 33 #define pof pop_front
 34 #define pob pop_back
 35 
 36 inline int Getint()
 37 {
 38     char Ch = ' ';
 39     int Ret = 0;
 40     while(!(Ch >= '0' && Ch <= '9')) Ch = getchar();
 41     while(Ch >= '0' && Ch <= '9')
 42     {
 43         Ret = Ret * 10 + Ch - '0';
 44         Ch = getchar();
 45     }
 46     return Ret;
 47 }
 48 
 49 const DB Eps = 1e-7;
 50 const string Name[] = {
 51     "Alex", 
 52     "Dmitry", 
 53     "Petr"};
 54 struct Point
 55 {
 56     DB x, y;
 57     
 58     inline void Read()
 59     {
 60         scanf("%lf%lf", &x, &y);
 61     }
 62 } A, B, C;
 63 DB D, V, W;
 64 DB Start, Cost;
 65 DB Ans[3];
 66 
 67 inline void Input()
 68 {
 69     A.Read();
 70     B.Read();
 71     C.Read();
 72     
 73     scanf("%lf%lf", &D, &V);
 74     D /= 60.0;
 75     
 76     int x = Getint();
 77     Start = x;
 78     x = Getint();
 79     Start += x / 60.0;
 80     x = Getint();
 81     Cost = x;
 82     x = Getint();
 83     Cost += x / 60.0;
 84     
 85     scanf("%lf", &W);
 86 }
 87 
 88 inline DB Sqr(DB x)
 89 {
 90     return x * x;
 91 }
 92 
 93 inline DB Dist(const Point &A, const Point &B)
 94 {
 95     return sqrt(Sqr(A.x - B.x) + Sqr(A.y - B.y));
 96 }
 97 
 98 inline DB Multi(const Point &O, const Point &A, const Point &B)
 99 {
100     DB  X1 = A.x - O.x,
101         X2 = B.x - O.x,
102         Y1 = A.y - O.y,
103         Y2 = B.y - O.y;
104     return X1 * Y2 - X2 * Y1;
105 }
106 
107 inline void Solve()
108 {
109     Ans[0] = Ans[2] = 9.0;
110     Ans[1] = Start + Cost;
111     
112     DB DAB = Dist(A, B);
113     Ans[2] += DAB / W;
114     if(fabs(Multi(A, B, C)) <= Eps) Ans[2] += D;
115     
116     Ans[0] += D;
117     DB DAC = Dist(A, C), DCB = Dist(C, B);
118     Ans[0] += (DAC + DCB) / V;
119     
120     int p = 0;
121     For(i, 1, 2)
122         if(Ans[i] < Ans[p]) p = i;
123     //Rep(i, 3) printf("%.3lf ", Ans[i]);
124     cout << Name[p] << endl;
125 }
126 
127 int main() {
128     //freopen("F.in", "r", stdin);
129      Input();
130      Solve();
131 return 0;
132 }
View Code

 

posted @ 2015-12-21 15:25  yanzx6  阅读(430)  评论(0)  编辑  收藏  举报