标签：acm

Osu!

Osu! is a very popular music game. Basically, it is a game about clicking. Some points will appear on the screen at some time, and you have to click them at a correct time.


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 t_i at place (x_i, y_i), and you should click it exactly at t_i at (x_i, y_i). 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 t_i and t_i+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.

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
54.5893762558

Hint
In memory of the best osu! player ever Cookiezi.
 

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;
}

[ACM] HDU 5078 Osu!

标签：acm

原文地址：http://blog.csdn.net/sr_19930829/article/details/40392997