| Time Limit: 2000MS | Memory Limit: 65536K | |||
| Total Submissions: 3743 | Accepted: 1150 | 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 xi and yi each — coordinates of the i-th corner (−1000 ≤ xi, yi ≤ 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 |
Hint
Source
给你两个圆，半径相等，求放在一个凸多边形里两个圆不碰到边界的圆心。两个圆是可以重叠的。
半平面交，向内推进R，然后求组成多边形的最远的两个点即可。
1 /* *********************************************** 2 Author :kuangbin 3 Created Time :2013/8/18 15:52:28 4 File Name :F:2013ACM练习专题学习计算几何半平面交POJ3384.cpp 5 ************************************************ */ 6 7 #include <stdio.h> 8 #include <string.h> 9 #include <iostream> 10 #include <algorithm> 11 #include <vector> 12 #include <queue> 13 #include <set> 14 #include <map> 15 #include <string> 16 #include <math.h> 17 #include <stdlib.h> 18 #include <time.h> 19 using namespace std; 20 const double eps = 1e-8; 21 const double PI = acos(-1.0); 22 int sgn(double x) 23 { 24 if(fabs(x) < eps) return 0; 25 if(x < 0) return -1; 26 else return 1; 27 } 28 struct Point 29 { 30 double x,y; 31 Point(){} 32 Point(double _x,double _y) 33 { 34 x = _x; y = _y; 35 } 36 Point operator -(const Point &b)const 37 { 38 return Point(x - b.x, y - b.y); 39 } 40 double operator ^(const Point &b)const 41 { 42 return x*b.y - y*b.x; 43 } 44 double operator *(const Point &b)const 45 { 46 return x*b.x + y*b.y; 47 } 48 }; 49 struct Line 50 { 51 Point s,e; 52 double k; 53 Line(){} 54 Line(Point _s,Point _e) 55 { 56 s = _s; e = _e; 57 k = atan2(e.y - s.y,e.x - s.x); 58 } 59 Point operator &(const Line &b)const 60 { 61 Point res = s; 62 double t = ((s - b.s)^(b.s - b.e))/((s - e)^(b.s - b.e)); 63 res.x += (e.x - s.x)*t; 64 res.y += (e.y - s.y)*t; 65 return res; 66 } 67 }; 68 //半平面交，直线的左边代表有效区域 69 bool HPIcmp(Line a,Line b) 70 { 71 if(fabs(a.k - b.k) > eps)return a.k < b.k; 72 return ((a.s - b.s)^(b.e - b.s)) < 0; 73 } 74 Line Q[1010]; 75 void HPI(Line line[], int n, Point res[], int &resn) 76 { 77 int tot = n; 78 sort(line,line+n,HPIcmp); 79 tot = 1; 80 for(int i = 1;i < n;i++) 81 if(fabs(line[i].k - line[i-1].k) > eps) 82 line[tot++] = line[i]; 83 int head = 0, tail = 1; 84 Q[0] = line[0]; 85 Q[1] = line[1]; 86 resn = 0; 87 for(int i = 2; i < tot; i++) 88 { 89 if(fabs((Q[tail].e-Q[tail].s)^(Q[tail-1].e-Q[tail-1].s)) < eps || fabs((Q[head].e-Q[head].s)^(Q[head+1].e-Q[head+1].s)) < eps) 90 return; 91 while(head < tail && (((Q[tail]&Q[tail-1]) - line[i].s)^(line[i].e-line[i].s)) > eps) 92 tail--; 93 while(head < tail && (((Q[head]&Q[head+1]) - line[i].s)^(line[i].e-line[i].s)) > eps) 94 head++; 95 Q[++tail] = line[i]; 96 } 97 while(head < tail && (((Q[tail]&Q[tail-1]) - Q[head].s)^(Q[head].e-Q[head].s)) > eps) 98 tail--; 99 while(head < tail && (((Q[head]&Q[head-1]) - Q[tail].s)^(Q[tail].e-Q[tail].e)) > eps) 100 head++; 101 if(tail <= head + 1)return; 102 for(int i = head; i < tail; i++) 103 res[resn++] = Q[i]&Q[i+1]; 104 if(head < tail - 1) 105 res[resn++] = Q[head]&Q[tail]; 106 } 107 Point p[1010]; 108 Line line[1010]; 109 //*两点间距离 110 double dist(Point a,Point b) 111 { 112 return sqrt((a-b)*(a-b)); 113 } 114 void change(Point a,Point b,Point &c,Point &d,double p)//将线段ab往左移动距离p 115 { 116 double len = dist(a,b); 117 double dx = (a.y - b.y)*p/len; 118 double dy = (b.x - a.x)*p/len; 119 c.x = a.x + dx; c.y = a.y + dy; 120 d.x = b.x + dx; d.y = b.y + dy; 121 } 122 int main() 123 { 124 //freopen("in.txt","r",stdin); 125 //freopen("out.txt","w",stdout); 126 int n; 127 double r; 128 while(scanf("%d%lf",&n,&r) == 2) 129 { 130 for(int i = 0;i < n;i++) 131 scanf("%lf%lf",&p[i].x,&p[i].y); 132 reverse(p,p+n); 133 for(int i = 0;i < n;i++) 134 { 135 Point t1,t2; 136 change(p[i],p[(i+1)%n],t1,t2,r); 137 line[i] = Line(t1,t2); 138 } 139 int resn; 140 HPI(line,n,p,resn); 141 int res1 = 0, res2 = 0; 142 for(int i = 0;i < resn;i++) 143 for(int j = i;j < resn;j++) 144 if(dist(p[i],p[j]) > dist(p[res1],p[res2])) 145 res1 = i, res2 = j; 146 printf("%.5f %.5f %.5f %.5f ",p[res1].x,p[res1].y,p[res2].x,p[res2].y); 147 } 148 return 0; 149 }