标签：hdu   最水题   入门   

题目链接：hdu 5078 Osu!

题面：

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.
 

题意：就求最大值，开始还以为贪心，直接暴力就好了。

代码：

#include <iostream>
#include <iomanip>
#include <cmath>
using namespace std;
long long int squared_dis(long long int x1,long long int y3,long long int x2,long long int y2)
{
	return (x1-x2)*(x1-x2)+(y3-y2)*(y3-y2);
}
long long int store_x[1005],store_y[1005],t[1005];
int main()
{
	int tt,n;
	cin>>tt;
	while(tt--)
	{
		double maxx=0;
	  cin>>n;
	  double tmp;
	  for(int i=0;i<n;i++)
	  {
  	    cin>>t[i]>>store_x[i]>>store_y[i];	
  	  }	
  	  for(int i=0;i<n-1;i++)
  	  {
  	  	for(int j=i+1;j<n;j++)
  	  	{
  	  		if(t[i]!=t[j])
  	  		tmp=sqrt(1.0*squared_dis(store_x[i],store_y[i],store_x[j],store_y[j]))/abs(t[i]-t[j]);
	  	  	if(tmp>maxx)
	  	  	maxx=tmp;
	     }
  	  }
  	  cout<<fixed<<setprecision(10)<<maxx<<endl;
	}
	return 0;
} 

HDU 5078 Osu!

标签：hdu   最水题   入门   

原文地址：http://blog.csdn.net/david_jett/article/details/45652071