poj 3608 Bridge Across Islands题解(凸包+求两凸包间的最近点对)

http://poj.org/problem?id=3608

#include<iostream>#include<fstream>#include<iomanip>#include<cstdio>#include<cstring>#include<algorithm>#include<cstdlib>#include<cmath>#include<set>#include<map>#include<queue>#include<stack>#include<string>#include<vector>#include<sstream>#include<ctime>#include<cassert>#define LL long long#define eps 1e-8#define inf 999999.0#define zero(a) abs(a)<eps#define N 20#define pi acos(-1.0)using namespace std;struct Point{    double x,y;}p[10005],q[10005],pos;vector<Point >s1,s2;double dist(Point p1,Point p2){    return sqrt((p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y));}double xmul(Point p0,Point p1,Point p2){    return (p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y);}//Graham_scan求凸包逆时针排序顶点bool cmp(Point p1,Point p2){    if(xmul(pos,p1,p2)>eps)  return true;    else if(zero(xmul(pos,p1,p2))&&dist(pos,p1)<dist(pos,p2))  return true;    return false;}void Graham_scan(vector<Point>&s,Point p[],int n){    for(int i=1;i<n;i++)        if(p[i].y<p[0].y||(zero(p[i].y - p[0].y)&&p[i].x<p[0].x))            swap(p[i],p[0]);    pos=p[0];    sort(p+1,p+n,cmp);    s.clear();    s.push_back(p[0]);s.push_back(p[1]);    int i;      for(i=2;i<n;i++){//求上凸壳           while(s.size()>1&&xmul(s[s.size()-2],s[s.size()-1],p[i])<eps){//这样构造的凸包是逆时针的。              s.pop_back();        }          s.push_back(p[i]);    }      int tmp=s.size();      s.push_back(p[n-2]);    for(i=n-3;i>=0;i--){//求下凸壳           while(s.size()>tmp&&xmul(s[s.size()-2],s[s.size()-1],p[i])<eps){              s.pop_back();          }          s.push_back(p[i]);       }}//得到向量a1b1和a2b2的位置关系double Get_angle(Point a1,Point b1,Point a2,Point b2){    Point t;    t.x=a2.x-(b2.x-a1.x);    t.y=a2.y-(b2.y-a1.y);    return xmul(a1,b1,t);}double Dist_Point_Seg(Point p,Point a,Point b){    Point t=p;    t.x+=a.y-b.y;t.y+=b.x-a.x;    if(xmul(a,t,p)*xmul(b,t,p)>eps)        return dist(p,a)+eps<dist(p,b)?dist(p,a):dist(p,b);    else        return fabs(xmul(p,a,b))/dist(a,b);}double Min_Dist_Two_Polygons(vector<Point>s1,vector<Point>s2){    int na=s1.size()-1,nb=s2.size()-1;    int lp=0,lq=0;    for(int i=1;i<na;i++)        if(s1[i].y<s1[lp].y||(zero(s1[i].y-s1[lp].y)&&s1[i].x<s1[lp].x))           lp=i;    for(int i=1;i<nb;i++)        if(s2[i].y>s2[lq].y||(zero(s2[i].y-s2[lq].y)&&s2[i].x>s2[lq].x))           lq=i;    double ans=dist(s1[lp],s2[lq]);    int tp=lp,tq=lq;    do{        double angle=Get_angle(s1[tp],s1[tp+1],s2[tq],s2[tq+1]);        //和两个凸包的边都重合        if(zero(angle)){            ans=min(ans,Dist_Point_Seg(s1[tp],s2[tq],s2[tq+1]));            ans=min(ans,Dist_Point_Seg(s1[tp+1],s2[tq],s2[tq+1]));            ans=min(ans,Dist_Point_Seg(s2[tq],s1[tp],s1[tp+1]));            ans=min(ans,Dist_Point_Seg(s2[tq+1],s1[tp],s1[tp+1]));            tp=(tp+1)%na;tq=(tq+1)%nb;        }        else{            //和第二个凸包的边重合            if(angle<-eps){                ans=min(ans,Dist_Point_Seg(s1[tp],s2[tq],s2[tq+1]));                tq=(tq+1)%nb;            }            //和第一个凸包的边重合            else{                ans=min(ans,Dist_Point_Seg(s2[tq],s1[tp],s1[tp+1]));                tp=(tp+1)%na;            }        }    }while(!(lp==tp&&lq==tq));    return ans;}int main(){    int n,m;    while(scanf("%d%d",&n,&m)!=EOF&&n+m){        for(int i=0;i<n;i++)            scanf("%lf%lf",&p[i].x,&p[i].y);        Graham_scan(s1,p,n);        for(int i=0;i<m;i++)            scanf("%lf%lf",&q[i].x,&q[i].y);        Graham_scan(s2,q,m);        printf("%.5f\n",Min_Dist_Two_Polygons(s1,s2));    }    return 0;}