HDU 5078.Osu(2014 鞍山赛区现场赛 I)

436

题目



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 ≤ 10 6), x i, and y i (0 ≤ x i, y i ≤ 10 6) 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  


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


Sample Output  


9.2195444573
54.5893762558



题解


两个 note 之间的距离与时间差的比值叫做难度
求整首歌的最大难度

直接按时间排序算一遍,求最大值即可
尽量多输出小数保证精度

代码


/*
By:OhYee
Github:OhYee
Blog:http://www.oyohyee.com/
Email:oyohyee@oyohyee.com

かしこいかわいい?
エリーチカ!
要写出来Хорошо的代码哦~
*/
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <set>
#include <list>
#include <queue>
#include <stack>
#include <string>
#include <vector>
#include <bitset>
#include <functional>

using namespace std;

const int INF = 0x7FFFFFFF;
const double eps = 1e-10;

const int maxn = 1005;

struct Node {
    double t,x,y;
    bool operator < (const Node &rhs)const {
        return t < rhs.t;
    }
    double operator - (const Node &rhs)const {
        return sqrt((x - rhs.x)*(x - rhs.x) + (y - rhs.y)*(y-rhs.y));
    }
};

Node note[maxn];

void Do() {
    int n;
    cin >> n;
    for(int i = 0;i < n;i++)
        cin >> note[i].t >> note[i].x >> note[i].y;
    sort(note,note + n);

    double Max = 0;
    for(int i = 1;i < n;i++)
        Max = max(Max,(note[i] - note[i - 1]) / (note[i].t - note[i - 1].t));

    cout << fixed << setprecision(20) << Max << endl;
}

int main() {
    cin.tie(0);
    cin.sync_with_stdio(false);

    int T;
    cin >> T;
    while(T--)
        Do();

    return 0;
}
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