HDU 6097 Mindis【几何】

Mindis

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2800    Accepted Submission(s): 563
Special Judge

Problem Description

The center coordinate of the circle C is O, the coordinate of O is (0,0) , and the radius is r.
P and Q are two points not outside the circle, and PO = QO.
You need to find a point D on the circle, which makes PD+QD minimum.
Output minimum distance sum.

 

Input

The first line of the input gives the number of test cases T; T test cases follow.
Each case begins with one line with r : the radius of the circle C.
Next two line each line contains two integers x , y denotes the coordinate of P and Q.

Limits
T≤500000
−100≤x,y≤100
1≤r≤100

 

Output

For each case output one line denotes the answer.
The answer will be checked correct if its absolute or relative error doesn't exceed 10−6.
Formally, let your answer be a, and the jury's answer be b. Your answer is considered correct if |a−b|max(1,b)≤10−6.

 

Sample Input

444 00 440 33 040 22 040 11 0

 

Sample Output

5.65685435.65685435.89450306.7359174

 

Source

2017 Multi-University Training Contest - Team 6

 

思路：作P、Q的反演点，即OP∗OP′=r2，OQ∗OQ′=r2，根据P′Q′与圆是否相交来判断D的位置，若相交则结果为P′Q′乘上比例OP/r，若不相交，则先求出P′Q′的中点D′，在根据比例确定D的位置，计算PD+QD即可。

#include<iostream>#include<cstdio>#include<cstring>#include<string>#include<cmath>#include<queue>#include<stack>#include<vector>#include<map>#include<set>#include<algorithm>using namespace std;#define ll long long#define ms(a,b)  memset(a,b,sizeof(a))#define maxn 510const int M=1e3+10;const int MM=2e3+10;const int inf=0x3f3f3f3f;const int mod=1e9+7;;const double eps=1e-8;struct node{    double x,y;}p[10];int dcmp(double x){    if(fabs(x)<=eps)return 0;    return x<0?-1:1;}double cal(double x1,double y1,double x2,double y2){    return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));}int main(){    int t;    scanf("%d",&t);    while(t--){        double r;        scanf("%lf",&r);        scanf("%lf%lf",&p[1].x,&p[1].y);        scanf("%lf%lf",&p[2].x,&p[2].y);        double op=cal(p[1].x,p[1].y,0,0);        if(dcmp(op)==0){            printf("%.7f\n",2*r);            continue;        }        double k=r*r/op/op;        p[3].x=p[1].x*k;p[3].y=p[1].y*k;        p[4].x=p[2].x*k;p[4].y=p[2].y*k;        p[5].x=(p[3].x+p[4].x)/2;        p[5].y=(p[4].y+p[3].y)/2;        double od=cal(p[5].x,p[5].y,0,0);        if(od<=r){            printf("%.7f\n",cal(p[3].x,p[3].y,p[4].x,p[4].y)*op/r);        }        else {            p[6].x=p[5].x*r/od;p[6].y=p[5].y*r/od;            printf("%.7f\n",2*cal(p[6].x,p[6].y,p[1].x,p[1].y));        }    }    return 0;}

Mindis

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2800    Accepted Submission(s): 563
Special Judge

Problem Description

The center coordinate of the circle C is O, the coordinate of O is (0,0) , and the radius is r.
P and Q are two points not outside the circle, and PO = QO.
You need to find a point D on the circle, which makes PD+QD minimum.
Output minimum distance sum.

 

Input

The first line of the input gives the number of test cases T; T test cases follow.
Each case begins with one line with r : the radius of the circle C.
Next two line each line contains two integers x , y denotes the coordinate of P and Q.

Limits
T≤500000
−100≤x,y≤100
1≤r≤100

 

Output

For each case output one line denotes the answer.
The answer will be checked correct if its absolute or relative error doesn't exceed 10−6.
Formally, let your answer be a, and the jury's answer be b. Your answer is considered correct if |a−b|max(1,b)≤10−6.

 

Sample Input

444 00 440 33 040 22 040 11 0

 

Sample Output

5.65685435.65685435.89450306.7359174

 

Source

2017 Multi-University Training Contest - Team 6