POJ 3608 旋转卡壳

思路：

旋转卡壳应用

注意点&边  边&边  点&点 三种情况

//By SiriusRen
#include <cmath>
#include <cstdio>
#include <algorithm>
using namespace std;
const double eps=1e-5;
const int N=10050;
typedef double db;
int n,m;
struct P{db x,y;P(){}P(db X,db Y){x=X,y=Y;}}p1[N],p2[N];
P operator-(P a,P b){return P(a.x-b.x,a.y-b.y);}
db operator*(P a,P b){return a.x*b.y-a.y*b.x;}
db operator^(P a,P b){return a.x*b.x+a.y*b.y;}
db dis(P c){return sqrt(c.x*c.x+c.y*c.y);}
db dotc(P a,P b,P c){return (b-a)^(c-a);}
db difc(P a,P b,P c){return (b-a)*(c-a);}
db p2seg(P a,P b,P c){
    if(dotc(a,b,c)<-eps)return dis(a-c);
    if(dotc(b,a,c)<-eps)return dis(b-c);
    return abs(difc(a,b,c)/dis(a-b));
}
db seg2seg(P a,P b,P c,P d){
    return min(min(p2seg(a,b,c),p2seg(a,b,d)),min(p2seg(c,d,a),p2seg(c,d,b)));
}
double solve(P p[],P q[],int np,int nq){
    int sp=1,sq=1;
    for(int i=1;i<=np;i++)if(p[i].y<p[sp].y)sp=i;
    for(int i=1;i<=nq;i++)if(q[i].y>q[sq].y)sq=i;
    p[np+1]=p[1],q[nq+1]=q[1];
    double tmp,ans=1e9;
    for(int i=1;i<=np;i++){
        while(tmp=(difc(p[sp+1],q[sq+1],p[sp])-difc(p[sp+1],q[sq],p[sp]))>eps)sq=sq%nq+1;
        if(tmp<-eps)ans=min(ans,p2seg(p[sp],p[sp+1],q[sq]));
        else ans=min(ans,seg2seg(p[sp],p[sp+1],q[sq],q[sq+1]));
        sp=sp%np+1;
    }return ans;
}
int main(){
    while(scanf("%d%d",&n,&m)&&(n||m)){
        for(int i=1;i<=n;i++)scanf("%lf%lf",&p1[i].x,&p1[i].y);
        for(int i=1;i<=m;i++)scanf("%lf%lf",&p2[i].x,&p2[i].y);
        printf("%lf\n",min(solve(p1,p2,n,m),solve(p2,p1,m,n)));
    }
}