题目链接：https://vjudge.net/problem/POJ-3608

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cmath>
 4 #include<algorithm>
 5 using namespace std;
 6 #define eps 1e-10
 7 const int N = 1e4+9;
 8 int n,m;
 9 struct Point{
10     double x,y;
11     Point operator - (const Point& b)const{
12         return (Point){x-b.x,y-b.y};
13     }
14     double operator ^ (const Point& b)const{
15         return x*b.y-b.x*y;
16     }
17     double operator * (const Point& b)const{
18         return x*b.x + y*b.y;
19     }
20 }pN[N],pM[N],p0;
21 bool cmp(Point a,Point b){
22     if(atan2(a.y-p0.y,a.x-p0.x) != atan2(b.y-p0.y,b.x-p0.x)){
23         return atan2(a.y-p0.y,a.x-p0.x) < atan2(b.y-p0.y,b.x-p0.x);
24     }
25     return a.x < b.x;
26 }
27 double dis(Point a,Point b){
28     return sqrt((a.x-b.x)*(a.x-b.x) + (a.y-b.y)*(a.y-b.y));
29 }
30 //点到线段最短距离 
31 inline double point_to_seg(Point a,Point s,Point t){
32     if(dis(s,t) < eps) return dis(a,s);
33     if( ((t-s)*(a-s)) < -eps) return dis(a,s);
34     if( ((s-t)*(a-t)) < -eps) return dis(a,t);
35     return fabs( ((t-s)^(a-s)) / dis(s,t) );
36 }
37 inline double seg_to_seg(Point A, Point B, Point C, Point D){
38     return min(min(point_to_seg(C,A,B), point_to_seg(D,A, B)), min(point_to_seg(A,C, D), point_to_seg(B,C, D)));
39 }
40 inline void init(){
41     int k;
42     scanf("%lf %lf",&pN[0].x,&pN[0].y);
43     p0 = pN[0];
44     for(int i = 1;i<n;++i){
45         scanf("%lf %lf",&pN[i].x,&pN[i].y);
46         if(p0.y>pN[i].y || (p0.y==pN[i].y && p0.x>pN[i].x)){
47             p0 = pN[i];
48             k = i;
49         }
50     }
51     pN[k] = pN[0];
52     pN[0] = p0;
53     sort(pN+1,pN+n,cmp);
54     
55     
56     scanf("%lf %lf",&pM[0].x,&pM[0].y);
57     p0 = pM[0];
58     for(int i = 1;i<m;++i){
59         scanf("%lf %lf",&pM[i].x,&pM[i].y);
60         if(p0.y>pM[i].y || (p0.y==pM[i].y && p0.x>pM[i].x)){
61             p0 = pM[i];
62             k = i;
63         }
64     }
65     pM[k] = pM[0];
66     pM[0] = p0;
67     sort(pM+1,pM+m,cmp);
68 }
69 inline void solve(){
70     int yminN = 0,ymaxM = 0;
71     for(int i = 1;i<n;++i) if(pN[i].y < pN[yminN].y) yminN = i;
72     for(int i = 1;i<m;++i) if(pM[i].y > pM[ymaxM].y) ymaxM = i;
73     pN[n] = pN[0];
74     pM[m] = pM[0];
75     double tem,ans = 0x3f3f3f3f;
76     for(int i = 0;i<n;++i){
77         while(tem = ( (pM[ymaxM+1]-pN[yminN+1])^(pN[yminN]-pN[yminN+1]) ) - ( (pM[ymaxM] - pN[yminN+1])^(pN[yminN]-pN[yminN+1])) > eps){
78             ymaxM++;
79             if(ymaxM == m) ymaxM = 0;
80         }
81         ans = min(ans,seg_to_seg(pN[yminN],pN[yminN+1],pM[ymaxM],pM[ymaxM+1]));
82         yminN++;
83         if(yminN == n) yminN = 0;
84     }
85     printf("%.5f\n",ans);
86 }
87 int main(){
88     while(~scanf("%d %d",&n,&m) && (n+m) ){
89         init();
90         solve();
91     }
92     return 0;
93 }

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