标签：des style blog class code ext
| 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
经典题目
代码：
/* ***********************************************
Author :_rabbit
Created Time :2014/5/10 16:26:51
File Name :20.cpp
************************************************ */
#pragma comment(linker, "/STACK:102400000,102400000")
#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <stdlib.h>
#include <string.h>
#include <limits.h>
#include <string>
#include <time.h>
#include <math.h>
#include <queue>
#include <stack>
#include <set>
#include <map>
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)<eps)return 0;
return 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<b.x||(a.x==b.x&&a.y<b.y);
}
bool operator == (const Point &a,const Point &b){
return dcmp(a.x-b.x)==0&&dcmp(a.y-b.y)==0;
}
double Dot(Point a,Point b){
return a.x*b.x+a.y*b.y;
}
double Length(Point a){
return sqrt(Dot(a,a));
}
double Angle(Point a,Point b){
return acos(Dot(a,b)/Length(a)/Length(b));
}
double angle(Point a){
return atan2(a.y,a.x);
}
double Cross(Point a,Point b){
return a.x*b.y-a.y*b.x;
}
Point vecunit(Point a){
return a/Length(a);
}
Point Normal(Point a){
return Point(-a.y,a.x)/Length(a);
}
Point Rotate(Point a,double rad){
return Point(a.x*cos(rad)-a.y*sin(rad),a.x*sin(rad)+a.y*cos(rad));
}
double Area2(Point a,Point b,Point c){
return Length(Cross(b-a,c-a));
}
double DistanceToSegment(Point p, Point a, Point b) {
if(a == b) return Length(p-a);
Point v1 = b-a, v2 = p-a, v3 = p-b;
if(dcmp(Dot(v1, v2)) < 0) return Length(v2);
else if(dcmp(Dot(v1, v3)) > 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<Point> CH(vector<Point> p){
sort(p.begin(),p.end());
p.erase(unique(p.begin(),p.end()),p.end());
int n=p.size();
int m=0;
vector<Point> ch(n+1);
for(int i=0;i<n;i++){
while(m>1&&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<Point> ch1,vector<Point> ch2)
{
int q=0, p=0,n=ch1.size(),m=ch2.size();
for(int i=0;i<n;i++) if(ch1[i].y-ch1[p].y < -eps) p=i;
for(int i=0;i<m;i++)if(ch2[i].y-ch2[q].y > eps) q=i;
ch1.push_back(ch1[0]);ch2.push_back(ch2[0]);
double tmp, ans=1e100;
for(int i=0;i<n;i++)
{
while((tmp = Cross(ch1[p+1]-ch1[p], ch2[q+1]-ch1[p]) - Cross(ch1[p+1]-ch1[p], ch2[q]- ch1[p])) > 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<Point> 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;
}POJ 3608 两凸包最近距离 旋转卡壳,布布扣,bubuko.com
标签：des style blog class code ext
原文地址：http://blog.csdn.net/xianxingwuguan1/article/details/25496525
