poj3608Bridge Across Islands+两凸包最近距离

Description

Thousands of thousands years ago there was a small kingdom located in the middle of the Pacific Ocean. The territory of the kingdom consists two separated islands. Due to the impact of the ocean current, the shapes of both the islands became convex polygons. The king of the kingdom wanted to establish a bridge to connect the two islands. To minimize the cost, the king asked you, the bishop, to find the minimal distance between the boundaries of the two islands.

Input

The input consists of several test cases.
Each test case begins with two integers N, M. (3 ≤ N, M ≤ 10000)
Each of the next N lines contains a pair of coordinates, which describes the position of a vertex in one convex polygon.
Each of the next M lines contains a pair of coordinates, which describes the position of a vertex in the other convex polygon.
A line with N = M = 0 indicates the end of input.
The coordinates are within the range [-10000, 10000].

Output

For each test case output the minimal distance. An error within 0.001 is acceptable.

Sample Input

4 4
0.00000 0.00000
0.00000 1.00000
1.00000 1.00000
1.00000 0.00000
2.00000 0.00000
2.00000 1.00000
3.00000 1.00000
3.00000 0.00000
0 0

Sample Output

1.00000

#include<cstdio>#include<cmath>#include<cstring>#include<iostream>#include<algorithm>#include<cstdlib>#include<queue>#include<map>#include<stack>#include<set>#define e exp(1.0); //2.718281828#define mod 1000000007#define INF 0x7fffffff#define inf 0x3f3f3f3ftypedef long long LL;using namespace std;#define zero(x) (((x)>0?(x):(-x))<eps)const double eps=1e-8;const double pi=acos(-1.0);//判断数k的符号 -1负数 1正数 0零int dcmp(double k) {    return k<-eps?-1:k>eps?1:0;}inline double sqr(double x) {    return x*x;}struct point {    double x,y;    point() {};    point(double a,double b):x(a),y(b) {};    void input() {        scanf("%lf %lf",&x,&y);    }    friend point operator + (const point &a,const point &b) {        return point(a.x+b.x,a.y+b.y);    }    friend point operator - (const point &a,const point &b) {        return point(a.x-b.x,a.y-b.y);    }    friend bool operator == (const point &a,const point &b) {        return dcmp(a.x-b.x)==0&&dcmp(a.y-b.y)==0;    }    friend point operator * (const point &a,const double &b) {        return point(a.x*b,a.y*b);    }    friend point operator * (const double &a,const point &b) {        return point(a*b.x,a*b.y);    }    friend point operator / (const point &a,const double &b) {        return point(a.x/b,a.y/b);    }    friend bool operator < (const point &a, const point &b) {        return a.x < b.x || (a.x == b.x && a.y < b.y);    }    double norm() {        return sqrt(sqr(x)+sqr(y));    }};//计算两个向量的叉积double cross(const point &a,const point &b) {    return a.x*b.y-a.y*b.x;}double cross3(point A,point B,point C) { //叉乘    return (B.x-A.x)*(C.y-A.y)-(B.y-A.y)*(C.x-A.x);}//计算两个点的点积double dot(const point &a,const point &b) {    return a.x*b.x+a.y*b.y;}double dot3(point A,point B,point C) { //点乘    return (C.x-A.x)*(B.x-A.x)+(C.y-A.y)*(B.y-A.y);}//向量长度double length(const point &a) {    return sqrt(dot(a,a));}//两个向量的角度double angle(const point &a,const point &b) {    return acos(dot(a,b)/length(a)/length(b));}//计算两个点的距离double dist(const point &a,const point &b) {    return (a-b).norm();}//op沿远点逆时针旋转角度Apoint rotate_point(const point &p,double A) {    double tx=p.x,ty=p.y;    return point(tx*cos(A)-ty*sin(A),tx*sin(A)+ty*cos(A));}double TriArea(const point &a, const point &b, const point &c) {    return fabs( cross( b - a, c - a ) ) / 2;}point Normal(const point &a) {    double L = length(a);    return point(-a.y/L, a.x/L);}//求两条直线的交点，p和q分别为两条直线上的点，v和w分别为直线的方向向量point GetLineIntersection(point p, point v, point q, point w) {    point u = p - q;    double t = cross(w, u) / cross(v, w);    return p + v * t;}//求点p到直线ab的距离double DistanceToLine(point p, point a, point b) {    point v1 = b - a, v2 = p - a;    return fabs(cross(v1,v2)) / length(v1);}//求点p到线段ab的距离double DistanceToSegment(point p, point a, point b) {    if(a==b) return length(p - a);    point v1 = b - a, v2 = p - a, v3 = p - b;    if(dcmp(dot(v1,v2)) < 0) return length(v2);    else if(dcmp(dot(v1,v3)) > 0) return length(v3);    else return fabs(cross(v1,v2)) / length(v1);}//判断直线a1a2和直线b1b2是否规范相交bool SegmentProperIntersection(point a1, point a2, point b1, point b2) {    double c1 = cross(a2-a1,b1-a1), c2 = cross(a2-a1, b2-a1);    double c3 = cross(b2-b1, a1-b1), c4 = cross(b2-b1, a2-b1);    return dcmp(c1) * dcmp(c2) <0 && dcmp(c3) * dcmp(c4) < 0;}//判断点p是否在直线a1a2上bool OnSegment(point p, point a1, point a2) {    return dcmp(cross(a1-p,a2-p)) ==0 && dcmp(dot(a1-p,a2-p))<0;}//判断线段a1a2和线段b1b2是否相交，可以在端点处相交bool SegmentIntersection(point a1, point a2, point b1, point b2) {    return SegmentProperIntersection(a1, a2, b1, b2) || OnSegment(a1, b1, b2) || OnSegment(a2, b1, b2);}double SegmentToSegment(point a1, point a2, point b1, point b2) {    //线段间的最短距离分为四种情况    double t1 = DistanceToSegment(b1, a1, a2);    double t2 = DistanceToSegment(b2, a1, a2);    double t3 = DistanceToSegment(a1, b1, b2);    double t4 = DistanceToSegment(a2, b1, b2);    return min(t1,min(t2,min(t3,t4)));}//使点集逆时针转void antiClockSort(point *ch, int n) {    double res = cross(ch[1] - ch[0], ch[2] - ch[0]);    if(dcmp(res) >= 0) return;    reverse(ch, ch+n);}int ConvexHull(point* P, int cnt, point* res) {    sort(P, P + cnt);    cnt = unique(P, P + cnt) - P;    int m = 0;    for (int i = 0; i < cnt; i++) {        while (m > 1 && cross(res[m - 1] - res[m - 2], P[i] - res[m - 2]) <= 0)            m--;        res[m++] = P[i];    }    int k = m;    for (int i = cnt - 2; i >= 0; i--) {        while (m > k && cross(res[m - 1] - res[m - 2], P[i] - res[m - 2]) <= 0)            m--;        res[m++] = P[i];    }    if (cnt > 1) m--;    return m;}//旋转卡壳求两个凸包最近距离double solve(point *P, point *Q, int n, int m) {    if(n==1 && m==1) {        return length(P[0] - Q[0]);    } else if(n==1 && m==2) {        return DistanceToSegment(P[0], Q[0], Q[1]);    } else if(n==2 && m==1) {        return DistanceToSegment(Q[0], P[0], P[1]);    } else if(n==2 && m==2) {        return SegmentToSegment(P[0], P[1], Q[0], Q[1]);    }    int yminP = 0, ymaxQ = 0;    for(int i=0; i<n; ++i) if(P[i].y < P[yminP].y) yminP = i;    for(int i=0; i<m; ++i) if(Q[i].y > Q[ymaxQ].y) ymaxQ = i;    P[n] = P[0];    Q[n] = Q[0];    double INF2 = 1e100;    double arg, ans = INF2;    for(int i=0; i<n; ++i) {        //当叉积负正转正时，说明点ymaxQ就是对踵点        while((arg=cross(P[yminP] - P[yminP+1],Q[ymaxQ+1] - Q[ymaxQ])) < -eps)            ymaxQ = (ymaxQ+1)%m;        double ret;        if(arg > eps) { //卡住第二个凸包上的点。            ret = DistanceToSegment(Q[ymaxQ], P[yminP], P[yminP+1]);            ans  = min(ans,ret);        } else { //arg==0，卡住第二个凸包的边            ret = SegmentToSegment(P[yminP],P[yminP+1],Q[ymaxQ],Q[ymaxQ+1]);            ans = min(ans,ret);        }        yminP = (yminP+1)%n;    }    return ans;}const int N=10005;point a[N],b[N];point cha[N],chb[N];int main() {    int n,m;    while(scanf("%d%d",&n,&m),n+m){        for(int i=0;i<n;++i)            scanf("%lf%lf",&a[i].x,&a[i].y);        for(int i=0;i<m;++i)            scanf("%lf%lf",&b[i].x,&b[i].y);        //使凸包的点逆时针        antiClockSort(a,n);        antiClockSort(b,m);        printf("%.5f\n",min(solve(a, b, n, m),solve(b,a,m,n)));    }    return 0;}