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POJ 3608 兩凸包最近距離旋轉卡殼

編輯：C++入門知識

Bridge Across Islands Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 8071 Accepted: 2364 Special Judge

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

經典題目

代碼：

/* ***********************************************
Author :_rabbit
Created Time :2014/5/10 16:26:51
File Name :20.cpp
************************************************ */
#pragma comment(linker, "/STACK:102400000,102400000")
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
#define INF 0x3f3f3f3f
#define eps 1e-5
#define pi acos(-1.0)
typedef long long ll;
int dcmp(double x){
	if(fabs(x)0?1:-1;
}
struct Point{
	double x,y;
	Point(double _x=0,double _y=0){
		x=_x;y=_y;
	}
};
Point operator + (const Point &a,const Point &b){  
    return Point(a.x+b.x,a.y+b.y);  
}  
Point operator - (const Point &a,const Point &b){  
    return Point(a.x-b.x,a.y-b.y);  
}  
Point operator * (const Point &a,const double &p){  
    return Point(a.x*p,a.y*p);  
}  
Point operator / (const Point &a,const double &p){  
    return Point(a.x/p,a.y/p);  
}  
bool operator < (const Point &a,const Point &b){  
    return a.x 0) return Length(v3);  
    else return fabs(Cross(v1, v2)) / Length(v1);  
}  
double dis_pair_seg(Point p1, Point p2, Point p3, Point p4)  {  
    return min(min(DistanceToSegment(p1, p3, p4), DistanceToSegment(p2, p3, p4)),  
     min(DistanceToSegment(p3, p1, p2), DistanceToSegment(p4, p1, p2)));  
}  
vector CH(vector p){
	sort(p.begin(),p.end());
	p.erase(unique(p.begin(),p.end()),p.end());
	int n=p.size();
	int m=0;
	vector ch(n+1);
	for(int i=0;i1&&Cross(ch[m-1]-ch[m-2],p[i]-ch[m-1])<=0)m--;
		ch[m++]=p[i];
	}
	int k=m;
	for(int i=n-2;i>=0;i--){
		while(m>k&&Cross(ch[m-1]-ch[m-2],p[i]-ch[m-2])<=0)m--;
		ch[m++]=p[i];
	}
	if(n>1)m--;
	ch.resize(m);
	return ch;
}
double RC_Distance(vector ch1,vector ch2)  
{  
    int q=0, p=0,n=ch1.size(),m=ch2.size();
    for(int i=0;i eps) q=i;  
    ch1.push_back(ch1[0]);ch2.push_back(ch2[0]); 
  
    double tmp, ans=1e100;  
    for(int i=0;i eps)  
            q=(q+1)%m;  
        if(tmp < -eps) ans = min(ans,DistanceToSegment(ch2[q],ch1[p],ch1[p+1]));  
        else ans = min(ans,dis_pair_seg(ch1[p],ch1[p+1],ch2[q],ch2[q+1]));  
        p=(p+1)%n;  
    }  
    return ans;  
}  

int main()
{
     //freopen("data.in","r",stdin);
     //freopen("data.out","w",stdout);
     int n,m;
	 while(cin>>n>>m){
		 if(n==0&&m==0)break;
		 vector ch1,ch2;
		 Point p;
		 while(n--)scanf("%lf%lf",&p.x,&p.y),ch1.push_back(p);
		 while(m--)scanf("%lf%lf",&p.x,&p.y),ch2.push_back(p);
		 ch1=CH(ch1);ch2=CH(ch2);
		 printf("%.5lf\n",RC_Distance(ch1,ch2));
	 }
     return 0;
}