23333我以前一定是学了假的旋转卡壳，，，，（以前只会面积的2333）
换姿势，这个题用两条平行线搞，一共那么有一个点和一条线，两条线平行，这些个情况，那么就维护出这些东西就好了。
（说的好简单啊2333）
1 #include <cstdio> 2 #include <iostream> 3 #include <cmath> 4 #define N 100005 5 #define LL long long 6 #define eps 1e-8 7 using namespace std; 8 inline int ra() 9 { 10 int x=0,f=1; char ch=getchar(); 11 while (ch<'0' || ch>'9') {if (ch=='-') f=-1; ch=getchar();} 12 while (ch>='0' && ch<='9') {x=x*10+ch-'0'; ch=getchar();} 13 return x*f; 14 } 15 struct point 16 { 17 double x,y; 18 }A[N],B[N]; 19 const double inf=1e50; 20 int n,m; 21 double cross(point c, point a, point b) 22 { 23 return (a.x-c.x)*(b.y-c.y)-(a.y-c.y)*(b.x-c.x); 24 } 25 double dot(point c, point a, point b) 26 { 27 return (a.x-c.x)*(b.x-c.x)+(a.y-c.y)*(b.y-c.y); 28 } 29 double dis(point a, point b) 30 { 31 return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y)); 32 } 33 double point_line_dis(point a, point b, point aim) 34 { 35 if (dis(a,b)<eps) return dis(aim,a); 36 if (dot(a,aim,b)<eps) return dis(aim,a); 37 if (dot(b,aim,a)<eps) return dis(aim,b); 38 return fabs(cross(aim,a,b))/dis(a,b); 39 } 40 double line_line_dis(point a, point b, point c, point d) 41 { 42 double dis1=min(point_line_dis(a,b,c),point_line_dis(a,b,d)); 43 double dis2=min(point_line_dis(c,d,a),point_line_dis(c,d,b)); 44 return min(dis1,dis2); 45 } 46 double RC(point *p, point *q, int n, int m) 47 { 48 p[n]=p[0]; q[m]=q[0]; 49 int down=0,up=0; 50 for (int i=0; i<n; i++) if (p[i].y<p[down].y) down=i; 51 for (int i=0; i<m; i++) if (q[i].y>q[up].y) up=i; 52 double ans=inf; 53 for (int i=0; i<n; i++) 54 { 55 double cnt; 56 while (cnt=cross(p[down+1],q[up+1],p[down])-cross(p[down+1],q[up],p[down])>eps) up=(up+1)%m; 57 if (cnt+eps<0) ans=min(ans,point_line_dis(p[down],p[down+1],q[up])); 58 else ans=min(ans,line_line_dis(p[down],p[down+1],q[up],q[up+1])); 59 down=(down+1)%n; 60 } 61 return ans; 62 } 63 int main(int argc, char const *argv[]) 64 { 65 while (scanf("%d%d",&n,&m) && (n||m)) 66 { 67 for (int i=0; i<n; i++) scanf("%lf%lf",&A[i].x,&A[i].y); 68 for (int i=0; i<m; i++) scanf("%lf%lf",&B[i].x,&B[i].y); 69 printf("%.5lf ",min(RC(A,B,n,m),RC(B,A,m,n))); 70 } 71 return 0; 72 }