poj 3384 半平面交

Feng Shui

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 5183		Accepted: 1548		Special Judge

Description

Feng shui is the ancient Chinese practice of placement and arrangement of space to achieve harmony with the environment. George has recently got interested in it, and now wants to apply it to his home and bring harmony to it.

There is a practice which says that bare floor is bad for living area since spiritual energy drains through it, so George purchased two similar round-shaped carpets (feng shui says that straight lines and sharp corners must be avoided). Unfortunately, he is unable to cover the floor entirely since the room has shape of a convex polygon. But he still wants to minimize the uncovered area by selecting the best placing for his carpets, and asks you to help.

You need to place two carpets in the room so that the total area covered by both carpets is maximal possible. The carpets may overlap, but they may not be cut or folded (including cutting or folding along the floor border) — feng shui tells to avoid straight lines.

Input

The first line of the input file contains two integer numbers n and r — the number of corners in George’s room (3 ≤ n ≤ 100) and the radius of the carpets (1 ≤ r ≤ 1000, both carpets have the same radius). The following n lines contain two integers x_i and y_i each — coordinates of the i-th corner (−1000 ≤ x_i, y_i ≤ 1000). Coordinates of all corners are different, and adjacent walls of the room are not collinear. The corners are listed in clockwise order.

Output

Write four numbers x₁, y₁, x₂, y₂ to the output file, where (x₁, y₁) and (x₂, y₂) denote the spots where carpet centers should be placed. Coordinates must be precise up to 4 digits after the decimal point.

If there are multiple optimal placements available, return any of them. The input data guarantees that at least one solution exists.

Sample Input

#1	5 2 -2 0 -5 3 0 8 7 3 5 0
#2	4 3 0 0 0 8 10 8 10 0

Sample Output

#1	-2 3 3 2.5
#2	3 5 7 3

/*
poj 3384

给你一个多边形,然后往里面放两个圆,问能够占的最大面积.
即相当于在能够放下两个圆的情况下使圆形离得尽可能的远.

如果将多边形的每条边往里面平移半径的长度R.那么剩下的多边形就是圆心的取点范围
然后在这个基础上枚举点即可.

主要是通过剩下的这些线段构成一个新的多边形,然后枚举它的顶点即可。
但是平移后的多边形的边的长度莫有变诶,所以可以看成多条直线围成的多边形。
那么就成了半平面交问题(即给你直线和图形在这条直线的哪一边,然后构成的多边形)

hhh-2016-05-17 19:55:56
*/
#include <iostream>
#include <vector>
#include <cstring>
#include <string>
#include <cstdio>
#include <queue>
#include <cmath>
#include <algorithm>
#include <functional>
#include <map>
using namespace std;
#define lson  (i<<1)
#define rson  ((i<<1)|1)
typedef long long ll;
using namespace std;
const int  maxn = 300;
const double PI = 3.1415926;
const double eps = 1e-10;

int sgn(double x)
{
    if(fabs(x) < eps) return 0;
    if(x < 0)
        return -1;
    else
        return 1;
}

struct Point
{
    double x,y;
    Point() {}
    Point(double _x,double _y)
    {
        x = _x,y = _y;
    }
    Point operator -(const Point &b)const
    {
        return Point(x-b.x,y-b.y);
    }
    double operator ^(const Point &b)const
    {
        return x*b.y-y*b.x;
    }
    double operator *(const Point &b)const
    {
        return x*b.x + y*b.y;
    }
};

struct Line
{
    Point s,t;
    double k;
    Line() {}
    Line(Point _s,Point _t)
    {
        s = _s;
        t = _t;
        k = atan2(t.y-s.y,t.x-s.x);
    }
    Point operator &(const Line &b) const
    {
        Point res = s;
        double ta = ((s-b.s)^(b.s-b.t))/((s-t)^(b.s-b.t));
        res.x += (t.x-s.x)*ta;
        res.y += (t.y-s.y)*ta;
        return res;
    }
};

bool HPIcmp(Line a,Line b)
{
    if(fabs(a.k-b.k) > eps) return a.k<b.k;
    return ((a.s-b.s)^(b.t-b.s)) < 0;
}
Line li[maxn];

double CalArea(Point p[],int n)
{
    double ans = 0;
    for(int i = 0; i < n; i++)
    {
        ans += (p[i]^p[(i+1)%n])/2;
    }
    return ans;
}

double dist(Point a,Point b)
{
    return sqrt((a-b)*(a-b));
}

void HPI(Line line[],int n,Point res[],int &resn)
{
    int tot =n;
    sort(line,line+n,HPIcmp);
    tot = 1;
    for(int i = 1; i < n; i++)
    {
        if(fabs(line[i].k - line[i-1].k) > eps)
            line[tot++] = line[i];
    }
    int head = 0,tail = 1;
    li[0] = line[0];
    li[1] = line[1];
    resn = 0;
    for(int i = 2; i < tot; i++)
    {
        if(fabs((li[tail].t-li[tail].s)^(li[tail-1].t-li[tail-1].s)) < eps||
                fabs((li[head].t-li[head].s)^(li[head+1].t-li[head+1].s)) < eps)
            return;
        while(head < tail && (((li[tail] & li[tail-1]) - line[i].s) ^ (line[i].t-line[i].s)) > eps)
            tail--;
        while(head < tail && (((li[head] & li[head+1]) - line[i].s) ^ (line[i].t-line[i].s)) > eps)
            head++;
        li[++tail] = line[i];
    }
    while(head < tail && (((li[tail] & li[tail-1]) - li[head].s) ^ (li[head].t-li[head].s)) > eps)
        tail--;
    while(head < tail && (((li[head] & li[head-1]) - li[tail].s) ^ (li[tail].t-li[tail].t)) > eps)
        head++;
    if(tail <= head+1)
        return;
    for(int i = head; i < tail; i++)
        res[resn++] = li[i]&li[i+1];
    if(head < tail-1)
        res[resn++] = li[head]&li[tail];
    double Mans = -1;
    int ansi,ansj;
    for(int i = 0; i < resn; i++)
    {
        for(int j = 0; j < resn; j++)
            if( sgn(dist(res[i],res[j])-Mans) >= 0)
            {
                Mans = dist(res[i],res[j]) ;
                ansi = i,ansj = j;
            }
    }
    printf("%.4f %.4f %.4f %.4f\n",res[ansi].x,res[ansi].y,res[ansj].x,res[ansj].y);
}
Point p0;
Point lis[maxn];
Line line[maxn];
bool cmp(Point a,Point b)
{
    double t = (a-p0)^(b-p0);
    if(sgn(t) > 0)return true;
    else if(sgn(t) == 0 && sgn(dist(a,lis[0])-dist(b,lis[0])) <= 0)
        return true;
    else
        return false;
}

Point ta,tb;
Point tans[maxn];
void fin(Point a,Point b,double mid)
{
    double len = dist(a,b);
    double dx = (a.y-b.y)*mid/len;
    double dy = (b.x-a.x)*mid/len;
    ta.x = a.x+dx,ta.y = a.y+dy;
    tb.x = b.x+dx,tb.y = b.y+dy;
}

int main()
{
    //freopen("in.txt","r",stdin);
    int n;
    double  r;
    while(scanf("%d%lf",&n,&r)!= EOF)
    {
        for(int i = 0; i < n; i++)
        {
            scanf("%lf%lf",&lis[i].x,&lis[i].y);
        }
        reverse(lis,lis+n);
        for(int i = 0; i < n; i++)
        {
            fin(lis[i],lis[(i+1)%n],r);
            line[i] = Line(ta,tb);
        }
        int resn;
        HPI(line,n,tans,resn);
    }
    return 0;
}