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 4 4 4 0 0 4 4 0 3 3 0 4 0 2 2 0 4 0 1 1 0   Sample Output 5.6568543 5.6568543 5.8945030 6.7359174   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 510
const 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 4 4 4 0 0 4 4 0 3 3 0 4 0 2 2 0 4 0 1 1 0   Sample Output 5.6568543 5.6568543 5.8945030 6.7359174   Source 2017 Multi-University Training Contest - Team 6