1 #include <iostream>  2 #include<cstdio>  3 #include<cstring>  4 #include<algorithm>  5 #include<stdlib.h>  6 #include<vector>  7 #include<cmath>  8 #include<queue>  9 #include<set> 10 using namespace std; 11 #define N 2010 12 #define LL long long 13 #define INF 0xfffffff 14 const double eps = 1e-8; 15 const double pi = acos(-1.0); 16 const double inf = ~0u>>2; 17 const int MAXN=1550; 18 int m; 19 double r; 20 int cCnt,curCnt;//此时cCnt为最终切割得到的多边形的顶点数、暂存顶点个数 21 struct point 22 { 23     double x,y; 24     point(double x=0,double y=0):x(x),y(y) {} 25 }; 26 typedef point pointt; 27 pointt operator -(point a,point b) 28 { 29     return point(a.x-b.x,a.y-b.y); 30 } 31 point points[MAXN],p[MAXN],q[MAXN];//读入的多边形的顶点（顺时针）、p为存放最终切割得到的多边形顶点的数组、暂存核的顶点 32 void getline(point x,point y,double &a,double &b,double   &c) //两点x、y确定一条直线a、b、c为其系数 33 { 34     a = y.y - x.y; 35     b = x.x - y.x; 36     c = y.x * x.y - x.x * y.y; 37 } 38 void initial() 39 { 40     for(int i = 1; i <= m; ++i)p[i] = points[i]; 41     p[m+1] = p[1]; 42     p[0] = p[m]; 43     cCnt = m;//cCnt为最终切割得到的多边形的顶点数，将其初始化为多边形的顶点的个数 44 } 45 point intersect(point x,point y,double a,double b,double c) //求x、y形成的直线与已知直线a、b、c、的交点 46 { 47     double u = fabs(a * x.x + b * x.y + c); 48     double v = fabs(a * y.x + b * y.y + c); 49     point pt; 50     pt.x=(x.x * v + y.x * u) / (u + v); 51     pt.y=(x.y * v + y.y * u) / (u + v); 52     return  pt; 53 } 54 void cut(double a,double b ,double c) 55 { 56     curCnt = 0; 57     for(int i = 1; i <= cCnt; ++i) 58     { 59         if(a*p[i].x + b*p[i].y + c >= 0)q[++curCnt] = p[i];// c由于精度问题，可能会偏小，所以有些点本应在右侧而没在， 60         //故应该接着判断 61         else 62         { 63             if(a*p[i-1].x + b*p[i-1].y + c > 0) //如果p[i-1]在直线的右侧的话， 64             { 65                 //则将p[i],p[i-1]形成的直线与已知直线的交点作为核的一个顶点(这样的话，由于精度的问题，核的面积可能会有所减少) 66                 q[++curCnt] = intersect(p[i],p[i-1],a,b,c); 67             } 68             if(a*p[i+1].x + b*p[i+1].y + c > 0) //原理同上 69             { 70                 q[++curCnt] = intersect(p[i],p[i+1],a,b,c); 71             } 72         } 73     } 74     for(int i = 1; i <= curCnt; ++i)p[i] = q[i];//将q中暂存的核的顶点转移到p中 75     p[curCnt+1] = q[1]; 76     p[0] = p[curCnt]; 77     cCnt = curCnt; 78 } 79 double dis(point a) 80 { 81     return sqrt(a.x*a.x+a.y*a.y); 82 } 83 void solve(int r) 84 { 85     //注意：默认点是顺时针，如果题目不是顺时针，规整化方向 86     initial(); 87     for(int i = 1; i <= m; ++i) 88     { 89         point ta, tb, tt; 90         tt.x = points[i+1].y - points[i].y; 91         tt.y = points[i].x - points[i+1].x; 92         double k = r*1.0 / sqrt(tt.x * tt.x + tt.y * tt.y); 93         tt.x = tt.x * k; 94         tt.y = tt.y * k; 95         ta.x = points[i].x + tt.x; 96         ta.y = points[i].y + tt.y; 97         tb.x = points[i+1].x + tt.x; 98         tb.y = points[i+1].y + tt.y; 99         double a,b,c;100         getline(ta,tb,a,b,c);101         cut(a,b,c);102     }103     double ans = -1;104     point p1, p2;105     int i,j;106     for(i  = 1; i <= curCnt ; i++)107         for(j = i ; j<=curCnt ; j++)108         {109             if(ans<dis(p[i]-p[j]))110             {111                 ans = dis(p[i]-p[j]);112                 p1 = p[i];113                 p2 = p[j];114             }115         }116     printf("%.4f %.4f %.4f %.4f\n",p1.x,p1.y,p2.x,p2.y);117 }118 /*void GuiZhengHua(){119      //规整化方向，逆时针变顺时针，顺时针变逆时针120     for(int i = 1; i < (m+1)/2; i ++)121       swap(points[i], points[m-i]);122 }*/123 int main()124 {125     int r,i;126     while(scanf("%d%d",&m,&r)!=EOF)127     {128         for(i  = 1; i <= m ; i++)129             scanf("%lf%lf",&points[i].x,&points[i].y);130         points[m+1] = points[1];131         solve(r);132     }133     return 0;134 }