[ACM] HDU 5078 Osu!
Osu!

Now, you want to write an algorithm to estimate how diffecult a game is.
To simplify the things, in a game consisting of N points, point i will occur at time ti at place (xi, yi), and you should click it exactly at ti at (xi, yi). That means you should move your cursor
from point i to point i+1. This movement is called a jump, and the difficulty of a jump is just the distance between point i and point i+1 divided by the time between ti and ti+1. And the difficulty of a game is simply the difficulty
of the most difficult jump in the game.
Now, given a description of a game, please calculate its difficulty.
For each test case, the first line contains an integer N (2 ≤ N ≤ 1000) denoting the number of the points in the game. Then N lines follow, the i-th line consisting of 3 space-separated integers, ti(0 ≤ ti < ti+1 ≤ 106),
xi, and yi (0 ≤ xi, yi ≤ 106) as mentioned above.
Your answer will be considered correct if and only if its absolute or relative error is less than 1e-9.
2
5
2 1 9
3 7 2
5 9 0
6 6 3
7 6 0
10
11 35 67
23 2 29
29 58 22
30 67 69
36 56 93
62 42 11
67 73 29
68 19 21
72 37 84
82 24 98
9.2195444573
54.5893762558HintIn memory of the best osu! player ever Cookiezi.
解题思路:
水题,看懂题意,写代码就没问题。
代码:
#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <string.h>
#include <cmath>
#include <iomanip>
#include <vector>
#include <map>
#include <stack>
#include <queue>
using namespace std;
int n; struct Point
{
int x,y,t;
}point[1002]; double dis(Point a,Point b)
{
return sqrt((double)(a.x-b.x)*(a.x-b.x)+(double)(a.y-b.y)*(a.y-b.y));
} int main()
{
int t;cin>>t;
while(t--)
{
cin>>n;
cin>>point[1].t>>point[1].x>>point[1].y;
double ans=-1;
for(int i=2;i<=n;i++)
{
cin>>point[i].t>>point[i].x>>point[i].y;
double temp=dis(point[i],point[i-1])/(point[i].t-point[i-1].t);
if(ans<temp)
ans=temp;
}
cout<<setiosflags(ios::fixed)<<setprecision(9)<<ans<<endl;
}
return 0;
}
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