poj3525Most Distant Point from the Sea（半平面交）

  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 }