hdu5826

physics

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 817    Accepted Submission(s): 454

Problem Description

There are n balls on a smooth horizontal straight track. The track can be considered to be a number line. The balls can be considered to be particles with the same mass.

At the beginning, ball i is at position Xi. It has an initial velocity of Vi and is moving in direction Di.(Di∈−1,1)
Given a constant C. At any moment, ball its acceleration Ai and velocity Vi have the same direction, and magically satisfy the equation that Ai * Vi = C.
As there are multiple balls, they may collide with each other during the moving. We suppose all collisions are perfectly elastic collisions.

There are multiple queries. Each query consists of two integers t and k. our task is to find out the k-small velocity of all the balls t seconds after the beginning.

* Perfectly elastic collision : A perfectly elastic collision is defined as one in which there is no loss of kinetic energy in the collision.

 

Input

The first line contains an integer T, denoting the number of testcases.

For each testcase, the first line contains two integers n <= 10^5 and C <= 10^9.
n lines follow. The i-th of them contains three integers Vi, Xi, Di. Vi denotes the initial velocity of ball i. Xi denotes the initial position of ball i. Di denotes the direction ball i moves in. 

The next line contains an integer q <= 10^5, denoting the number of queries.
q lines follow. Each line contains two integers t <= 10^9 and 1<=k<=n.
1<=Vi<=10^5,1<=Xi<=10^9

 

Output

For each query, print a single line containing the answer with accuracy of 3 decimal digits.

 

Sample Input

13 73 3 13 10 -12 7 132 31 23 3

 

Sample Output

6.0834.796
7.141
思路:球碰撞之后速度不会变化，跟原先状态一样，然后可以找出t时刻的速度(2*c*t+v*v)的 开方，所以直接给初始速度排个序就可以了；
前面一直TLE，后面才发现数组开小了。
代码：
#include<iostream>#include<cmath>#include<algorithm>#include<cstdio>using namespace std;struct ball{long long int v;int p;int d;double v1;};bool cmp1(ball x,ball y){         return x.v<=y.v;}ball b1[111111];int main(){int T ;cin>>T;while(T--){int num,C;scanf("%d%d",&num,&C);int i,j,k;for(i=1;i<=num;i++){scanf("%lld%d%d",&b1[i].v,&b1[i].p,&b1[i].d);}int que;long long int time;int rank;cin>>que;sort(b1+1,b1+1+num,cmp1);for(i=1;i<=que;i++){scanf("%lld%d",&time,&rank);b1[rank].v1=sqrt(2*C*1.0*time+b1[rank].v*b1[rank].v*1.0);printf("%.3lf\n",b1[rank].v1);    }}return 0;}