CSU 2088: Pigs can't take a sudden turn(简单的计算几何)

xiaoxiao2021-02-28  3


Pigs can’t take a sudden turn

Time limit:1000 ms Memory limit:65536 kB

Problem Descrpition

You maybe have ACed a question about computational geometry,which describes two dogs’ journey and calculate the shortest distance between them. Now, I provide a easier problem about two pigs. heir path are half-lines. When they start running, they won’t stop. Why? Because they can’t take a sudden turn.

So, I provide the start points of two pigs and their velocities. Please tell me the shortest distance between them.

We purpose that two pigs start running at the same time and will not stop because they are running in an vast grasslane without any tree.

If they smash into each other, the answer is zero and ignore the influence.


The first line is an integer T which indicates the number of cases. For each case, first line is four numbers x1,y1,x2,y2, indicate the start points of two pigs(x1,y1), (x2,y2). The second line is four numbers u1,v1,u2,v2, indicate the velocities of two pigs(u1,v1),(u2,v2). T <= 1000 The absolute value of x1,y1,x2,y2,u1,v1,u2,v2 will not lager than 1000.


For each case, you should output one line like ”Case i: d”. i stands for the case number and d stands for the answer, which should rounded to 6 decimal places.

Sample Input

5 1 1 2 2 1 1 2 2 1 1 2 2 1 1 -1 -1 1 1 2 2 0 1 0 -1 1 1 1 1 1 1 2 2 0 0 0 1 0 1 1 0

Sample Output

Case 1: 1.414214 Case 2: 0.000000 Case 3: 1.000000 Case 4: 0.000000 Case 5: 0.707107






#include <cstdio> #include <iostream> #include <cmath> #define y1 YY using namespace std; const double eps=1e-8; double a,b,c; double x1,y1,u1,v1; double x2,y2,u2,v2; double len(double t) { double ans=a*t*t+b*t+c; return sqrt(ans); } int main() { int T,ca=1; scanf("%d",&T); while(T--) { scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2); scanf("%lf%lf%lf%lf",&u1,&v1,&u2,&v2); c=(x1-x2)*(x1-x2)+(y1-y2)*(y1-y2); b=2*(x1-x2)*(u1-u2)+2*(y1-y2)*(v1-v2); a=(u1-u2)*(u1-u2)+(v1-v2)*(v1-v2); printf("Case %d: ",ca++); if(fabs(a)<eps) { printf("%.6f\n",len(0)); continue; } if(-b/(a*2)<0) { printf("%.6f\n",len(0)); } else { printf("%.6f\n",len(-b/(a*2))); } } return 0; }
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