hdu 5078 Osu!（鞍山现场赛）

最新推荐文章于 2022-02-20 00:29:14 发布

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最新推荐文章于 2022-02-20 00:29:14 发布

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分类专栏： ACM-区域赛文章标签： algorithm

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本文深入探讨了Osu!游戏的难度计算算法，通过分析游戏内点位与时间戳，计算玩家跳跃操作的难度，并提供了一个高效实现此功能的代码实例。该文不仅适用于游戏开发者，也对算法爱好者具有一定的参考价值。

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Osu!

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 20 Accepted Submission(s): 15
Special Judge

Problem Description

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.

Input

The first line contains an integer T (T ≤ 10), denoting the number of the test cases.

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, t_i(0 ≤ t_i < t_i+1 ≤ 10⁶), x_i, and y_i (0 ≤ x_i, y_i ≤ 10⁶) as mentioned above.

Output

For each test case, output the answer in one line.

Your answer will be considered correct if and only if its absolute or relative error is less than 1e-9.

Sample Input

Sample Output

9.2195444573
54.5893762558

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

求最大难度，难度为相邻两点的距离除以时间差。

代码：

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <math.h>
using namespace std;
double a[10000];
double b[10000];
int ti[10000];
double dis(int i,int j)
{
    return sqrt((a[i]-a[j])*(a[i]-a[j])+(b[i]-b[j])*(b[i]-b[j]));
}
int main()
{
    int t,n;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        double ans=0;

        for(int i=0;i<n;i++)
        {
            scanf("%d%lf%lf",&ti[i],&a[i],&b[i]);
        }
        for(int i=1;i<n;i++)
        {
            ans=max(ans,(dis(i-1,i)/(ti[i]-ti[i-1])));
        }
        printf("%.10f\n",ans);
    }
    return 0;
}