2014ACM/ICPC亚洲区鞍山赛区现场赛1009Osu!
鞍山的签到题,求两点之间的距离除以时间的最大值。直接暴力过的。
Time Limit:1000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u
System Crawler (2014-10-22)
Description

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 iat (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
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
Your answer will be considered correct if and only if its absolute or relative error is less than 1e-9.
Sample Input
Sample Output
Hint
In memory of the best osu! player ever Cookiezi.
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<stdlib.h>
using namespace std;
const int maxn=+;
struct
{
double t;
double x;
double y
;
}Node[maxn];
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
int N;
double maxd=-;
scanf("%d",&N);
for(int i=;i<N;i++)
scanf("%lf%lf%lf",&Node[i].t,&Node[i].x,&Node[i].y);
for(int i=;i<N;i++)
{
for(int j=i+;j<N;j++)
{
double time=sqrt((Node[i].x-Node[j].x)*(Node[i].x-Node[j].x)+(Node[i].y-Node[j].y)*(Node[i].y-Node[j].y))/fabs(Node[i].t-Node[j].t);
maxd=max(maxd,time);
}
}
printf("%.10lf\n",maxd);
}
return ;
}
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