hdu 6097 Mindis

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Mindis

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1182    Accepted Submission(s): 135
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

 

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题意：

圆心 O 坐标(0, 0), 给定两点 P, Q（不在圆外），满足 PO = QO,

要在圆上找一点 D，使得 PD + QD 取到最小值。

此题会补充三种做法：

1.椭圆

第一定义

平面内与两定点

  

、

  

的距离的和等于常数

  

（

  

）的动点P的轨迹叫做椭圆，即：

第二定义

椭圆平面内到定点

  

（c,0）的距离和到定直线

  

：

  

（

  

不在

  

上）的距离之比为常数

  

（即离心率

  

，0<e<1）的点的轨迹是椭圆。

其中定点

  

为椭圆的焦点[5]  ，定直线

  

称为椭圆的准线（该定直线的方程是

  

（焦点在x轴上），或

  

（焦点在y轴上））。

其他定义

根据椭圆的一条重要性质：椭圆上的点与椭圆长轴（事实上只要是直径都可以）两端点连线的斜率之积是定值，定值为

  

（前提是长轴平行于x轴。若长轴平行于y轴，比如焦点在y轴上的椭圆，可以得到斜率之积为 -a²/b²=1/（e²-1）），可以得出：

在坐标轴内，动点（

  

）到两定点（

  

）（

  

）的斜率乘积等于常数m（-1<m<0）

注意：考虑到斜率不存在时不满足乘积为常数，所以

  

无法取到，即该定义仅为去掉两个点的椭圆。

令PQ为椭圆的焦点。那么PQ=2c c显然可以得到，对离心率e二分，直到直到满足条件的e，对于一个椭圆来说，椭圆上的点到两焦点的距离2a，所以如果我能找到一个合适的离心率构造出椭圆，最后答案就是2a，但我得满足这个椭圆与圆有交点，不然取不到圆上的点。c是固定值，如果我要改变e，a是未知值，我只能改变b，使离心率改变，所以就有了如下代码。

//china no.1#pragma comment(linker, "/STACK:1024000000,1024000000")#include <vector>#include <iostream>#include <string>#include <map>#include <stack>#include <cstring>#include <queue>#include <list>#include <stdio.h>#include <set>#include <algorithm>#include <cstdlib>#include <cmath>#include <iomanip>#include <cctype>#include <sstream>#include <functional>#include <stdlib.h>#include <time.h>#include <bitset>using namespace std;#define pi acos(-1)#define endl '\n'#define srand() srand(time(0));#define me(x,y) memset(x,y,sizeof(x));#define foreach(it,a) for(__typeof((a).begin()) it=(a).begin();it!=(a).end();it++)#define close() ios::sync_with_stdio(0); cin.tie(0);#define FOR(x,n,i) for(int i=x;i<=n;i++)#define FOr(x,n,i) for(int i=x;i<n;i++)#define W while#define sgn(x) ((x) < 0 ? -1 : (x) > 0)#define bug printf("***********\n");#define db doubletypedef long long LL;const int INF=0x3f3f3f3f;const LL LINF=0x3f3f3f3f3f3f3f3fLL;const int dx[]={-1,0,1,0,1,-1,-1,1};const int dy[]={0,1,0,-1,-1,1,-1,1};const int maxn=2e2+10;const int maxx=1e4+100;const double EPS=1e-7;const int mod=10000007;template<class T>inline T min(T a,T b,T c) { return min(min(a,b),c);}template<class T>inline T max(T a,T b,T c) { return max(max(a,b),c);}template<class T>inline T min(T a,T b,T c,T d) { return min(min(a,b),min(c,d));}template<class T>inline T max(T a,T b,T c,T d) { return max(max(a,b),max(c,d));}inline LL Scan(){    int f=1;char C=getchar();LL x=0;    while (C<'0'||C>'9'){if (C=='-')f=-f;C=getchar();}    while (C>='0'&&C<='9'){x=x*10+C-'0';C=getchar();}    x*=f;return x;}//freopen( "in.txt" , "r" , stdin );//freopen( "data.out" , "w" , stdout );//cerr << "run time is " << clock() << endl;struct Point{    double x, y;    Point(double x = 0, double y = 0) : x(x), y(y) { }    inline void input()    {        scanf("%lf%lf",&x,&y);    }    inline void print()    {        printf("%.6lf %.6lf\n",x,y);    }};db r;db dis(Point a,Point b){    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));}int megumi(db e,db c,db B,db r)//e为离心率{    db y=(1-e*e)/(e*e)*B;//椭圆上一点且该点为切点    db a=c/e;    db b=sqrt(a*a-c*c);    if(y>b) return (b+B>r);    double x2=a*a*(1-y*y/(b*b));//x^2    return (x2+(y+B)*(y+B)>r*r)||(B+b>r);//判断条件}int main(){    int t;    t=Scan();    while(t--)    {        scanf("%lf",&r);        Point a,b;        Point O={0,0};        a.input(),b.input();        double o1=dis(O,a),o2=dis(O,b),o12=dis(a,b);        if(fabs(o1-r)<1e-4||fabs(o2-r)<1e-4)//有一个点在圆上        {            printf("%.7f\n",o12);            continue;        }        else if(fabs(o12-o1-o2)<1e-4)//同在圆心        {            printf("%.7f\n",2*r);            continue;        }        else if(fabs(o12)<1e-4)//P Q重点        {            printf("%.7f\n",2*(r-o1));            continue;        }        db c=o12/2;        db B=sqrt(o1*o1-c*c);        db l=1e-9,rr=1;        //cout<<c<<" "<<B<<endl;        int len=35;        while(len--)        {            double m=(l+rr)/2;            if(megumi(m,c,B,r)) l=m;            else rr=m;        }        printf("%.7f\n",c/l*2);//2a    }}

2.黄金分割三分（不会，只有一份用map的）

为啥可以map呢，-100到100所以可以map

#include<cstdio>#include<iostream>#include<cstring>#include<string>#include<algorithm>#include<map>#include<cmath>#include<vector>#include<stack>#include<climits>#include<ctime>#include<queue>#define FILEIN freopen("in.txt", "r", stdin)#define FILEOUT freopen("out.txt", "w", stdout)#define CLOSEIO ios::sync_with_stdio(false)#define PI acos(-1)#define CLR(a) memset(a,0,sizeof(a))#define MEM(a,x) memset(a,x,sizeof(a))#define eps 1e-8#define sf(x) scanf("%d",&x)#define PB(x) push_back(x)#define MP(x, y) make_pair(x, y)#define lowbit(x) x&(-x)#define fi first#define se second#define rep(a,b,c) for(int (a)=(b);(a)<(c);(a)++)#define drep(a,b,c) for(int (a)=(b);(a)>(c);--(a))#define dbg(x) cout << #x << "=" << x << endl#define _ixvii0ivusing namespace std;const int maxn = 1e5+5;typedef long long ll;typedef double db;const int inf = INT_MAX;const ll INF = LLONG_MAX;const ll mod = 1e9 + 7;ll mul(ll x,ll y){return x*y%mod;}ll q_mul(ll a, ll b){ ll ans = 0;while(b){if(b & 1){ans=(ans+a)%mod;} b>>=1;a=(a+a) % mod;}return ans;}ll q_pow(ll x , ll y){ll res=1;while(y){if(y&1) res=q_mul(res,x) ; y>>=1 ; x=q_mul(x,x);} return res;}ll inv(ll x) { return q_pow(x, mod-2); }int Read() {    int x = 0, F = 1; char C = getchar();    while (C < '0' || C > '9') { if (C == '-') F = -F; C = getchar(); }    while (C >= '0' && C <= '9') { x = x * 10 - '0' + C, C = getchar(); }    return x * F;}struct vis{int r;int R;int len;bool operator<(const vis &a)const{if(r!=a.r)return r<a.r;else if(R!=a.R)return R<a.R;else return len<a.len;}bool operator=(const vis &a)const{return r==a.r&&R==a.R&&len==a.len;}};map<vis,db>mp;namespace fastIO{    #define BUF_SIZE 100000    #define OUT_SIZE 100000    #define ll long long    //fread->read    bool IOerror=0;    inline char nc(){        static char buf[BUF_SIZE],*p1=buf+BUF_SIZE,*pend=buf+BUF_SIZE;        if (p1==pend){            p1=buf; pend=buf+fread(buf,1,BUF_SIZE,stdin);            if (pend==p1){IOerror=1;return -1;}            //{printf("IO error!\n");system("pause");for (;;);exit(0);}        }        return *p1++;    }    inline bool blank(char ch){return ch==' '||ch=='\n'||ch=='\r'||ch=='\t';}    inline void read(int &x){        bool sign=0; char ch=nc(); x=0;        for (;blank(ch);ch=nc());        if (IOerror)return;        if (ch=='-')sign=1,ch=nc();        for (;ch>='0'&&ch<='9';ch=nc())x=x*10+ch-'0';        if (sign)x=-x;    }    inline void read(double &x){        bool sign=0; char ch=nc(); x=0;        for (;blank(ch);ch=nc());        if (IOerror)return;        if (ch=='-')sign=1,ch=nc();        for (;ch>='0'&&ch<='9';ch=nc())x=x*10+ch-'0';        if (ch=='.'){            double tmp=1; ch=nc();            for (;ch>='0'&&ch<='9';ch=nc())tmp/=10.0,x+=tmp*(ch-'0');        }        if (sign)x=-x;    }    inline void read(char *s){        char ch=nc();        for (;blank(ch);ch=nc());        if (IOerror)return;        for (;!blank(ch)&&!IOerror;ch=nc())*s++=ch;        *s=0;    }    inline void read(char &c){        for (c=nc();blank(c);c=nc());        if (IOerror){c=-1;return;}    }    //getchar->read    inline void read1(int &x){        char ch;int bo=0;x=0;        for (ch=getchar();ch<'0'||ch>'9';ch=getchar())if (ch=='-')bo=1;        for (;ch>='0'&&ch<='9';x=x*10+ch-'0',ch=getchar());        if (bo)x=-x;    }    inline void read1(ll &x){        char ch;int bo=0;x=0;        for (ch=getchar();ch<'0'||ch>'9';ch=getchar())if (ch=='-')bo=1;        for (;ch>='0'&&ch<='9';x=x*10+ch-'0',ch=getchar());        if (bo)x=-x;    }    inline void read1(double &x){        char ch;int bo=0;x=0;        for (ch=getchar();ch<'0'||ch>'9';ch=getchar())if (ch=='-')bo=1;        for (;ch>='0'&&ch<='9';x=x*10+ch-'0',ch=getchar());        if (ch=='.'){            double tmp=1;            for (ch=getchar();ch>='0'&&ch<='9';tmp/=10.0,x+=tmp*(ch-'0'),ch=getchar());        }        if (bo)x=-x;    }    inline void read1(char *s){        char ch=getchar();        for (;blank(ch);ch=getchar());        for (;!blank(ch);ch=getchar())*s++=ch;        *s=0;    }    inline void read1(char &c){for (c=getchar();blank(c);c=getchar());}    //scanf->read    inline void read2(int &x){scanf("%d",&x);}    inline void read2(ll &x){        #ifdef _WIN32            scanf("%I64d",&x);        #else        #ifdef __linux            scanf("%lld",&x);        #else            puts("error:can't recognize the system!");        #endif        #endif    }    inline void read2(double &x){scanf("%lf",&x);}    inline void read2(char *s){scanf("%s",s);}    inline void read2(char &c){scanf(" %c",&c);}    inline void readln2(char *s){gets(s);}    //fwrite->write    struct Ostream_fwrite{        char *buf,*p1,*pend;        Ostream_fwrite(){buf=new char[BUF_SIZE];p1=buf;pend=buf+BUF_SIZE;}        void out(char ch){            if (p1==pend){                fwrite(buf,1,BUF_SIZE,stdout);p1=buf;            }            *p1++=ch;        }        void print(int x){            static char s[15],*s1;s1=s;            if (!x)*s1++='0';if (x<0)out('-'),x=-x;            while(x)*s1++=x%10+'0',x/=10;            while(s1--!=s)out(*s1);        }        void println(int x){            static char s[15],*s1;s1=s;            if (!x)*s1++='0';if (x<0)out('-'),x=-x;            while(x)*s1++=x%10+'0',x/=10;            while(s1--!=s)out(*s1); out('\n');        }        void print(ll x){            static char s[25],*s1;s1=s;            if (!x)*s1++='0';if (x<0)out('-'),x=-x;            while(x)*s1++=x%10+'0',x/=10;            while(s1--!=s)out(*s1);        }        void println(ll x){            static char s[25],*s1;s1=s;            if (!x)*s1++='0';if (x<0)out('-'),x=-x;            while(x)*s1++=x%10+'0',x/=10;            while(s1--!=s)out(*s1); out('\n');        }        void print(double x,int y){            static ll mul[]={1,10,100,1000,10000,100000,1000000,10000000,100000000,                1000000000,10000000000LL,100000000000LL,1000000000000LL,10000000000000LL,                100000000000000LL,1000000000000000LL,10000000000000000LL,100000000000000000LL};            if (x<-1e-12)out('-'),x=-x;x*=mul[y];            ll x1=(ll)floor(x); if (x-floor(x)>=0.5)++x1;            ll x2=x1/mul[y],x3=x1-x2*mul[y]; print(x2);            if (y>0){out('.'); for (size_t i=1;i<y&&x3*mul[i]<mul[y];out('0'),++i); print(x3);}        }        void println(double x,int y){print(x,y);out('\n');}        void print(char *s){while (*s)out(*s++);}        void println(char *s){while (*s)out(*s++);out('\n');}        void flush(){if (p1!=buf){fwrite(buf,1,p1-buf,stdout);p1=buf;}}        ~Ostream_fwrite(){flush();}    }Ostream;    inline void print(int x){Ostream.print(x);}    inline void println(int x){Ostream.println(x);}    inline void print(char x){Ostream.out(x);}    inline void println(char x){Ostream.out(x);Ostream.out('\n');}    inline void print(ll x){Ostream.print(x);}    inline void println(ll x){Ostream.println(x);}    inline void print(double x,int y){Ostream.print(x,y);}    inline void println(double x,int y){Ostream.println(x,y);}    inline void print(char *s){Ostream.print(s);}    inline void println(char *s){Ostream.println(s);}    inline void println(){Ostream.out('\n');}    inline void flush(){Ostream.flush();}    //puts->write    char Out[OUT_SIZE],*o=Out;    inline void print1(int x){        static char buf[15];        char *p1=buf;if (!x)*p1++='0';if (x<0)*o++='-',x=-x;        while(x)*p1++=x%10+'0',x/=10;        while(p1--!=buf)*o++=*p1;    }    inline void println1(int x){print1(x);*o++='\n';}    inline void print1(ll x){        static char buf[25];        char *p1=buf;if (!x)*p1++='0';if (x<0)*o++='-',x=-x;        while(x)*p1++=x%10+'0',x/=10;        while(p1--!=buf)*o++=*p1;    }    inline void println1(ll x){print1(x);*o++='\n';}    inline void print1(char c){*o++=c;}    inline void println1(char c){*o++=c;*o++='\n';}    inline void print1(char *s){while (*s)*o++=*s++;}    inline void println1(char *s){print1(s);*o++='\n';}    inline void println1(){*o++='\n';}    inline void flush1(){if (o!=Out){if (*(o-1)=='\n')*--o=0;puts(Out);}}    struct puts_write{        ~puts_write(){flush1();}    }_puts;    inline void print2(int x){printf("%d",x);}    inline void println2(int x){printf("%d\n",x);}    inline void print2(char x){printf("%c",x);}    inline void println2(char x){printf("%c\n",x);}    inline void print2(ll x){        #ifdef _WIN32            printf("%I64d",x);        #else        #ifdef __linux            printf("%lld",x);        #else            puts("error:can't recognize the system!");        #endif        #endif    }    inline void println2(ll x){print2(x);printf("\n");}    inline void println2(){printf("\n");}    #undef ll    #undef OUT_SIZE    #undef BUF_SIZE};using namespace fastIO;struct point{    db x,y;    point() {};    point (db _x,db _y)    {        x = _x;        y = _y;    }    inline void input()    {        read(x);        read(y);    }};inline int sgn(db x){    return x<-eps?-1:x>eps?1:0;}typedef point line;inline db Dot(line a,line b){    return a.x*b.x+a.y*b.y;}inline line operator +(line a,line b){    return line(a.x+b.x,a.y+b.y);}inline line operator -(line a,line b){    return line(a.x-b.x,a.y-b.y);}inline line operator *(line b,db a){    return line(b.x*a,b.y*a);}inline db len(line a){    return sqrt(Dot(a,a));}inline db Dis(point a,point b){return len(b-a);}inline line Rotate(line a,db rad)//rad是弧度 负数表示顺时针 正数表示逆时针{db cr  =cos(rad);db sr = sin(rad);    return line(a.x*cr-a.y*sr,a.x*sr+a.y*cr);}struct circle{point c;double r;circle() {}circle(point _c,double _r){c = _c;r = _r;}inline void input(){c.input();read(r);}};typedef point line;int main(){//FILEIN;//FILEOUT;    int t = Read();    int ca = 1;    mp.clear();    while(t--){db R;point a,b;read(R);a.input();b.input();int rr = a.x*a.x+a.y*a.y;int RR = R*R;int lenn = (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y);//cout << rr << " " << RR <<" " <<lenn << endl;if(!(a.x-b.x)&&!(a.y-b.y)){println(2*(R-len(a)),7);continue;}if(!(a.x+b.x)&&!(a.y+b.y)||!a.x&&!a.y){println(2*R,7);continue;}vis vs;vs.R = RR;vs.r = rr;vs.len = lenn;if(mp[vs]){println(mp[vs],7);continue;}point mab;mab.x = (a.x+b.x)/2.0;mab.y = (a.y+b.y)/2.0;line mm = mab;db l = 0; db r = PI;int tm = 0;db x = R/len(mm);mm.x*=x;mm.y*=x;db diss;while(r-l>eps){if(tm>50) break;db mid1 = (l+r)/2.0;db mid2 = (r+mid1)/2.0;line l1 = Rotate(mm,mid1);line l2 = Rotate(mm,mid2);db dis1 = Dis(l1,a)+Dis(l1,b);db dis2 = Dis(l2,a)+Dis(l2,b);if(sgn(dis2-dis1)>0) r = mid2,diss = dis1;else l = mid1,diss = dis2;tm++;}mp[vs] = diss;println(diss,7);//puts("");    }}

3.反演

反演：

设在平面内给定一点O和常数k(k不等于零)，对于平面内任意一点A，确定A′，使A′在直线OA上一点，并且有向线段OA与OA′满足OA·OA′=k，我们称这种变换是以O为的反演中心，以k为反演幂的反演变换，简称反演。——百度百科

DP+DQ=(DP'+DQ')*k

当 P'Q' 与圆有交点时：

不妨设交点为 O'，若 D 不为 O'，则 P'D + Q'D >  P'Q'（三角形两边之和大于第三边）；当且仅当 D 取 O' 时，P'Q + Q'D 取到最小值，即为 P'Q'。

当 P'Q' 与圆无交点时：

 PQ 的中垂线与圆的交点一定是使得DP'+DQ'最小的点。OQ'=OP'，且DP'=DQ'，请仔细想想这个为什么。

//china no.1#pragma comment(linker, "/STACK:1024000000,1024000000")#include <vector>#include <iostream>#include <string>#include <map>#include <stack>#include <cstring>#include <queue>#include <list>#include <stdio.h>#include <set>#include <algorithm>#include <cstdlib>#include <cmath>#include <iomanip>#include <cctype>#include <sstream>#include <functional>#include <stdlib.h>#include <time.h>#include <bitset>using namespace std;#define pi acos(-1)#define endl '\n'#define srand() srand(time(0));#define me(x,y) memset(x,y,sizeof(x));#define foreach(it,a) for(__typeof((a).begin()) it=(a).begin();it!=(a).end();it++)#define close() ios::sync_with_stdio(0); cin.tie(0);#define FOR(x,n,i) for(int i=x;i<=n;i++)#define FOr(x,n,i) for(int i=x;i<n;i++)#define W while#define sgn(x) ((x) < 0 ? -1 : (x) > 0)#define bug printf("***********\n");#define db doubletypedef long long LL;const int INF=0x3f3f3f3f;const LL LINF=0x3f3f3f3f3f3f3f3fLL;const int dx[]={-1,0,1,0,1,-1,-1,1};const int dy[]={0,1,0,-1,-1,1,-1,1};const int maxn=2e2+10;const int maxx=1e4+100;const double EPS=1e-7;const int mod=10000007;template<class T>inline T min(T a,T b,T c) { return min(min(a,b),c);}template<class T>inline T max(T a,T b,T c) { return max(max(a,b),c);}template<class T>inline T min(T a,T b,T c,T d) { return min(min(a,b),min(c,d));}template<class T>inline T max(T a,T b,T c,T d) { return max(max(a,b),max(c,d));}inline LL Scan(){    int f=1;char C=getchar();LL x=0;    while (C<'0'||C>'9'){if (C=='-')f=-f;C=getchar();}    while (C>='0'&&C<='9'){x=x*10+C-'0';C=getchar();}    x*=f;return x;}//freopen( "in.txt" , "r" , stdin );//freopen( "data.out" , "w" , stdout );//cerr << "run time is " << clock() << endl;struct Point{    double x, y;    Point(double x = 0, double y = 0) : x(x), y(y) { }    inline void input()    {        scanf("%lf%lf",&x,&y);    }    inline void print()    {        printf("%.6lf %.6lf\n",x,y);    }};db r;db dis(Point a,Point b){    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));}typedef Point Vector;Vector operator + (Vector A, Vector B) { return Vector(A.x + B.x, A.y + B.y); }Vector operator - (Vector A, Vector B) { return Vector(A.x - B.x, A.y - B.y); }Vector operator * (Vector A, double p) { return Vector(A.x * p, A.y * p); }Vector operator / (Vector A, double p) { return Vector(A.x / p, A.y / p); }bool operator < (const Point& a, const Point b){    return a.x < b.x || (a.x == b.x && a.y < b.y);}int dcmp(double x){    if(fabs(x) < EPS) return 0;    else return x < 0 ? -1 : 1;}bool operator == (const Point& a, const Point& b){    return dcmp(a.x-b.x) == 0 && dcmp(a.y-b.y);}//向量点积double Dot(Vector A, Vector B) { return A.x*B.x + A.y*B.y; }//向量长度double Length(Vector A) { return sqrt(Dot(A, A)); }//向量叉积  |a||b|sin<a,b>//叉积的结果也是一个向量，是垂直于向量a，b所形成的平面，如果看成三维坐标的话是在 z 轴上，上面结果是它的模。double Cross(Vector A, Vector B) { return A.x*B.y - A.y*B.x; }//点到直线的距离double DistanceToLine(Point P, Point A, Point B){    Vector v1 = B - A, v2 = P - A;    return fabs(Cross(v1, v2) / Length(v1)); //不取绝对值，得到的是有向距离}int main(){    //freopen( "in.txt" , "r" , stdin );    //freopen( "out.txt" , "w" , stdout );    int t;    t=Scan();    while(t--)    {        scanf("%lf",&r);        Point a,b;        Point O={0,0};        a.input(),b.input();        double o1=dis(O,a),o2=dis(O,b),o12=dis(a,b);        if(fabs(o1-r)<1e-4||fabs(o2-r)<1e-4)//有一个点在圆上        {            printf("%.10f\n",o12);            continue;        }        else if(fabs(o12-o1-o2)<1e-4)//圆心重合        {            printf("%.10f\n",2*r);            continue;        }        else if(fabs(o12)<1e-4)//P Q重点        {            printf("%.10f\n",2*(r-o1));            continue;        }        db k=o1/r;        db o11=(db)r*r/o1;//op'        db t=o11/o1;        Point p1={t*a.x,t*a.y};        Point q1={t*b.x,t*b.y};        db len=DistanceToLine(O,p1,q1);        if(dcmp(len-r)<=0)//p1q1与圆相交        {            printf("%.10f\n",dis(p1,q1)*k);            continue;        }        else        {            db h=len-r;            db d=dis(p1,q1)/2;            db ans=2*sqrt(h*h+d*d);            printf("%.10f\n",ans*k);        }    }}