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 100000 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 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 27 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 28 { 29     a = y.y - x.y; 30     b = x.x - y.x; 31     c = y.x * x.y - x.x * y.y; 32 } 33 double dis(point a,point b) 34 { 35     return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y)); 36 } 37 void initial() 38 { 39     for(int i = 1; i <= m; ++i)p[i] = points[i]; 40     p[m+1] = p[1]; 41     p[0] = p[m]; 42     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 43 } 44 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 45 { 46     double u = fabs(a * x.x + b * x.y + c); 47     double v = fabs(a * y.x + b * y.y + c); 48     point pt; 49     pt.x=(x.x * v + y.x * u) / (u + v); 50     pt.y=(x.y * v + y.y * u) / (u + v); 51     return  pt; 52 } 53 void cut(double a,double b ,double c) 54 { 55     curCnt = 0; 56     for(int i = 1; i <= cCnt; ++i) 57     { 58         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， 59         //故应该接着判断 60         else 61         { 62             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， 63             { 64                 //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) 65                 q[++curCnt] = intersect(p[i],p[i-1],a,b,c); 66             } 67             if(a*p[i+1].x + b*p[i+1].y + c > 0) //原理同上 68             { 69                 q[++curCnt] = intersect(p[i],p[i+1],a,b,c); 70             } 71         } 72     } 73     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 74     p[curCnt+1] = q[1]; 75     p[0] = p[curCnt]; 76     cCnt = curCnt; 77 } 78 int solve(double r) 79 { 80     //注意：默认点是顺时针，如果题目不是顺时针，规整化方向 81     initial(); 82 //    for(int i = 1; i <= m; ++i) 83 //    { 84 //        double a,b,c; 85 //        getline(points[i],points[i+1],a,b,c); 86 //        cut(a,b,c); 87 //    } 88  89       //如果要向内推进r，用该部分代替上个函数 90       for(int i = 1; i <= m; ++i){ 91           point ta, tb, tt; 92           tt.x = points[i+1].y - points[i].y; 93           tt.y = points[i].x - points[i+1].x; 94           double k = r / sqrt(tt.x * tt.x + tt.y * tt.y); 95           tt.x = tt.x * k; 96           tt.y = tt.y * k; 97           ta.x = points[i].x + tt.x; 98           ta.y = points[i].y + tt.y; 99           tb.x = points[i+1].x + tt.x;100           tb.y = points[i+1].y + tt.y;101           double a,b,c;102           getline(ta,tb,a,b,c);103           cut(a,b,c);104       }105     //多边形核的面积106 //    double area = 0;107 //    for(int i = 1; i <= curCnt; ++i)108 //        area += p[i].x * p[i + 1].y - p[i + 1].x * p[i].y;109 //    area = fabs(area / 2.0);110 //    printf("%.2f\n",area);111     if(curCnt) return 1;112     return 0;113 114 }115 void GuiZhengHua(){116      //规整化方向，逆时针变顺时针，顺时针变逆时针117     for(int i = 1; i < (m+1)/2; i ++)118       swap(points[i], points[m-i]);119 }120 //void change(double d)121 //{122 //    int i;123 //    for(i = 1; i <= m ;i++)124 //    {125 //        double len = dis(p[i],points[i+1]);126 //        double a = points[i+1].y-points[i].y;127 //        double b = points[i].x-points[i+1].x;128 //        double cos = a/len;129 //        double sin = b/len;130 //        points[i] = point(points[i].x+cos*d,points[i].y+sin*d);131 //        points[i+1] = point(points[i+1].x+cos*d,points[i+1].y+sin*d);132 //    }133 //}134 int main()135 {136     int i;137     while(scanf("%d",&m)&&m)138     {139         for(i = 1 ; i<=m; i++)140         scanf("%lf%lf",&points[i].x,&points[i].y);141         GuiZhengHua();142         points[m+1] = points[1];143         double rig = INF,lef = 0,mid;144         while(rig-lef>eps)145         {146             mid = (rig+lef)/2.0;147             //change(mid);148             if(solve(mid))149             lef = mid;150             else rig = mid;151         }152         printf("%.6f\n",lef);153     }154     return 0;155 }