usaco Closed Fences

来源：互联网发布：淘宝车牌选号编辑：程序博客网时间：2024/05/16 16:14
本来这道题目却被不想写的，因为方法有点麻烦。
不过最后还是耐下性子写了。呜……
思路是挺简单的。
设观察位置为A，BC为多边形上的一条边。若有边挡住了A点的视线，刚更新A当前的视野。
如下图：
此时视野为BC1
此时视野为B1C
此时被完全挡住，没有视野。
代码如下：
/*ID: guo geerPROG: fence4LANG: C++*/#include<iostream>#include<cstdio>#include<cstring>#include<string>#include<algorithm>#include<cmath>using namespace std;#define les 1e-6struct Point{int x, y;};Point P[300],ob;int res[300];double x,y;//交点 double cross(double x1,double y1,double x2,double y2,double x,double y){       return (x1-x)*(y2-y)-(x2-x)*(y1-y);}int intersection(double x1,double y1,double x2,double y2,double x3,double y3,double x4,double y4){       if(cross(x2,y2,x3,y3,x1,y1)==0&&cross(x2,y2,x4,y4,x1,y1)==0)       return 0;       if(cross(x2,y2,x3,y3,x1,y1)*cross(x2,y2,x4,y4,x1,y1)<=les&&cross(x4,y4,x1,y1,x3,y3)*cross(x4,y4,x2,y2,x3,y3)<=les)       return 1;       return 0;}int isInTriangle(double x1,double y1,double x2,double y2,double x3,double y3,double x,double y){    double x0,y0;    x0=(x1+x2+x3)/3;    y0=(y1+y2+y3)/3;    if(cross(x1,y1,x0,y0,x2,y2)*cross(x1,y1,x,y,x2,y2)<=les) return 0;    if(cross(x2,y2,x0,y0,x3,y3)*cross(x2,y2,x,y,x3,y3)<=les) return 0;    if(cross(x3,y3,x0,y0,x1,y1)*cross(x3,y3,x,y,x1,y1)<=les) return 0;    return 1;}void PointOfIntersection(double x1,double y1,double x2,double y2,double x3,double y3,double x4,double y4){      double k1,k2,b1,b2;      if(x1==x2&&x3!=x4)      {            k2=(y3-y4)/(x3-x4);            b2=y3-k2*x3;            x = x1;            y = k2*x1+b2;      }      else if(x1!=x2&&x3==x4)      {            k1=(y2-y1)/(x2-x1);            b1=y1-k1*x1;            x = x3;            y = k1*x3+b1;      }      else if(x1!=x2&&x3!=x4)      {            k1=(y2-y1)/(x2-x1);            b1=y1-k1*x1;            k2=(y3-y4)/(x3-x4);            b2=y3-k2*x3;            x = (b1-b2)/(k2-k1);            y = k1*x+b1;      }}int main(){    freopen("fence4.in","r",stdin);    freopen("fence4.out","w",stdout);    int n,Count;    while(scanf("%d", &n) == 1)    {     memset(res, 0, sizeof(res));     Count = 0;         scanf("%d %d",&ob.x,&ob.y);         for(int i=0; i<n; i++)         scanf("%d %d",&P[i].x,&P[i].y);                  for(int i=0; i<n; i++)         {              if(cross(ob.x,ob.y,P[i].x,P[i].y,P[(i+1)%n].x,P[(i+1)%n].y)==0)              continue;                    //printf("--------%d-----------\n",i);              bool flag = true;              double ax,ay,bx,by;              ax=P[i].x,ay=P[i].y;              bx=P[(i+1)%n].x,by=P[(i+1)%n].y;              for(int j=(i+1)%n; j!=i; j=(j+1)%n)              {                     if(cross(ob.x,ob.y,P[j].x,P[j].y,P[(j+1)%n].x,P[(j+1)%n].y)==0)                     continue;                     if(intersection(P[j].x,P[j].y,P[(j+1)%n].x,P[(j+1)%n].y,ob.x,ob.y,ax,ay)==1)                     {                           if(intersection(ob.x,ob.y,bx,by,P[(j+1)%n].x,P[(j+1)%n].y,P[j].x,P[j].y)==1)                           {                                //if(P[i].x==5&&P[i].y==50)                                                                                                       //printf("j=%d 1\n",j);                                                                                                       flag = false;                                break;                           }                           else if(isInTriangle(ob.x,ob.y,ax,ay,bx,by,P[j].x,P[j].y)==1)                           {                                //if(P[i].x==5&&P[i].y==50)                                     //printf("j=%d 2\n",j);                                PointOfIntersection(ob.x,ob.y,P[j].x,P[j].y,ax,ay,bx,by);                                ax=x;                                ay=y;                           }                           else if(isInTriangle(ob.x,ob.y,ax,ay,bx,by,P[(j+1)%n].x,P[(j+1)%n].y)==1)                           {                                //if(P[i].x==5&&P[i].y==50)                                       //printf("j=%d 3\n",j);                                PointOfIntersection(ob.x,ob.y,P[(j+1)%n].x,P[(j+1)%n].y,ax,ay,bx,by);                                ax=x;                                ay=y;                           }                                                                                                   }                     else if(intersection(P[j].x,P[j].y,P[(j+1)%n].x,P[(j+1)%n].y,ob.x,ob.y,bx,by)==1)                     {                           if(isInTriangle(ob.x,ob.y,ax,ay,bx,by,P[j].x,P[j].y)==1)                           {                                //if(P[i].x==5&&P[i].y==50)                                     //printf("j=%d 4\n",j);                                                                                PointOfIntersection(ob.x,ob.y,P[j].x,P[j].y,ax,ay,bx,by);                                bx=x;                                by=y;                           }                           else if(isInTriangle(ob.x,ob.y,ax,ay,bx,by,P[(j+1)%n].x,P[(j+1)%n].y)==1)                           {                                //if(P[i].x==5&&P[i].y==50)                                      //printf("j=%d 5\n",j);                                PointOfIntersection(ob.x,ob.y,P[(j+1)%n].x,P[(j+1)%n].y,ax,ay,bx,by);                                bx=x;                                by=y;                           }                                   }              }              //printf("%d %d %d %d %lf %lf %lf %lf\n",P[i].x,P[i].y,P[i+1].x,P[i+1].y,ax,ay,bx,by);              if(ax==bx&&ay==by)              {                     continue;              }              if(flag==true)              {                     Count ++;                     res[i]=1;              }         }     printf("%d\n",Count); for(int i=0; i<n-2; i++) if(res[i]==1) { printf("%d %d %d %d\n",P[i].x,P[i].y,P[i+1].x,P[i+1].y); } if(res[n-1]==1) printf("%d %d %d %d\n",P[0].x,P[0].y,P[n-1].x,P[n-1].y); if(res[n-2]==1) printf("%d %d %d %d\n",P[n-2].x,P[n-2].y,P[n-1].x,P[n-1].y);    }    return 0;}