USACO6.5-Closed Fences：计算几何

  1 /*
  2 TASK:fence4
  3 LANG:C++
  4 */
  5 
  6 #include <iostream>
  7 #include <vector>
  8 #include <stdlib.h>
  9 #include <stdio.h>
 10 #include <cmath>
 11 using namespace std;
 12 
 13 //定义点
 14 struct Point
 15 {
 16     double x;
 17     double y;
 18 };
 19 typedef Point Vector;
 20 bool operator == (const Point& p1, const Point& p2)
 21 {
 22     return fabs(p1.x - p2.x)<1e-12 && fabs(p1.y - p2.y)<1e-12;
 23 }
 24 typedef struct Point point;
 25 Vector operator - (const Point& A, const Point& B)
 26 {
 27     return Vector{A.x-B.x, A.y-B.y};
 28 }
 29 double Cross(const Vector& A, const Vector& B)
 30 {
 31     return A.x*B.y - A.y*B.x;
 32 }
 33 //叉积
 34 double multi(point p0, point p1, point p2)
 35 {
 36     return (p1.x - p0.x )*( p2.y - p0.y )-( p2.x -p0.x )*( p1.y -p0.y );
 37 }
 38 //点积
 39 double Dot(Vector A,Vector B)
 40 {
 41     return A.x*B.x+A.y*B.y;
 42 }
 43 
 44 //相交返回true,否则为false, 接口为两线段的端点
 45 bool isIntersected(point s1,point e1, point s2,point e2)
 46 {
 47     if(s1==s2 || s1==e2 || e1==s2 || e1==e2)
 48     {
 49         return false;
 50     }
 51     return (max(s1.x,e1.x) >=min(s2.x,e2.x)) &&
 52            (max(s2.x,e2.x)>=min(s1.x,e1.x)) &&
 53            (max(s1.y,e1.y) >=min(s2.y,e2.y)) &&
 54            (max(s2.y,e2.y) >=min(s1.y,e1.y)) &&
 55            (multi(s1,s2,e1)*multi(s1,e1,e2)>0) &&
 56            (multi(s2,s1,e2)*multi(s2,e2,e1)>0);
 57 }
 58 
 59 point ps[300];
 60 double x,y;
 61 int n ;
 62 bool check(int idx)
 63 {
 64     point pos=point {x,y};
 65     // 特判pos与线段idx平行的情况
 66     if(fabs(Cross(ps[idx]-pos,ps[idx+1]-pos))<1e-13)
 67     {
 68         return false;
 69     }
 70 
 71     bool flag=false;
 72     for(int i=0; i<n; i++)
 73     {
 74         // 将ps[idx]移动一小段距离
 75         if(i==idx)
 76             continue;
 77         double dx=(ps[idx+1].x-ps[idx].x)*1e-8;
 78         double dy=(ps[idx+1].y-ps[idx].y)*1e-8;
 79 
 80         if(isIntersected(pos,point {ps[idx].x+dx,ps[idx].y+dy},ps[i],ps[i+1]))
 81             flag=true;
 82     }
 83     if(!flag)
 84         return true;
 85     for(int i=0; i<n; i++)
 86     {
 87         if(i==idx)
 88             continue;
 89 
 90 
 91         double dx=(ps[idx].x-ps[idx+1].x)*1e-8;
 92         double dy=(ps[idx].y-ps[idx+1].y)*1e-8;
 93 
 94         if(isIntersected(pos,point {ps[idx+1].x+dx,ps[idx+1].y+dy},ps[i],ps[i+1]))
 95             return false;
 96     }
 97     return true;
 98 }
 99 
100 
101 int main()
102 {
103     freopen("fence4.in","r",stdin);
104     freopen("fence4.out","w",stdout);
105 
106     cin>>n;
107 
108     scanf("%lf%lf",&x,&y);
109     for(int i=0; i<n; i++)
110     {
111         scanf("%lf%lf",&ps[i].x,&ps[i].y);
112     }
113     ps[n]=ps[0];
114 
115     // 判断是否符合
116     for(int i=1; i<n; i++)
117     {
118         for(int j=0; j<i-1; j++)
119         {
120             if(isIntersected(ps[j],ps[j+1],ps[i],ps[i+1]))
121             {
122                 puts("NOFENCE");
123                 exit(0);
124             }
125         }
126     }
127 
128     vector<int> ans;
129     for(int i=0; i<n; i++)
130     {
131         if(check(i))
132             ans.push_back(i);
133     }
134 
135     int sz=ans.size();
136     if(sz>=2 && ans[sz-1]==n-1 && ans[sz-2]==n-2)
137         swap(ans[sz-1],ans[sz-2]);
138     printf("%d\n",sz);
139     for(int i=0; i<sz; i++)
140     {
141         if(ans[i]==n-1)
142             printf("%.0f %.0f %.0f %.0f\n",ps[ans[i]+1].x,ps[ans[i]+1].y,ps[ans[i]].x,ps[ans[i]].y);
143         else
144             printf("%.0f %.0f %.0f %.0f\n",ps[ans[i]].x,ps[ans[i]].y,ps[ans[i]+1].x,ps[ans[i]+1].y);
145     }
146 
147     return 0;
148 }