POJ-1584-A Round Peg in a Ground Hole-计算几何-凸多边形+多边形包含圆
http://poj.org/problem?id=1584
题意：顺时针或逆时针的点，让你先判断多边形是否为凸多边形，如果不是输出HOLE IS ILL-FORMED
如果是，判断能不能把一个给定大小和位置的圆完全包含
if (ok)
printf("PEG WILL FIT\n");
else
printf("PEG WILL NOT FIT\n");
1先把点都变成逆时针，
然后判凸，
然后判圆心是否在多边形内，是的话再看是否被包含
#include <cstdio>
#include <cmath>
#include <cstring>
#include <string>
#include <algorithm>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <iostream>
using namespace std;
const double pi=acos(-1.0);
double eps=1e-5;
double tm[5][35];
double max(double a,double b)
{return a>b?a:b;}
double min(double a,double b)
{return a<b?a:b;}
struct POINT
{
double x;
double y;
POINT(double a=0, double b=0) { x=a; y=b;} //constructor
};
struct LINESEG
{
POINT s;
POINT e;
LINESEG(POINT a, POINT b) { s=a; e=b;}
LINESEG() { }
};
struct LINE // 直线的解析方程 a*x+b*y+c=0 为统一表示，约定 a >= 0
{
double a;
double b;
double c;
LINE(double d1=1, double d2=-1, double d3=0) {a=d1; b=d2; c=d3;}
};
double multiply(POINT sp,POINT ep,POINT op)
{
return((sp.x-op.x)*(ep.y-op.y)-(ep.x-op.x)*(sp.y-op.y));
}
double area_of_polygon(int vcount,POINT polygon[])
{
int i;
double s;
if (vcount<3)
return 0;
s=polygon[0].y*(polygon[vcount-1].x-polygon[1].x);
for (i=1;i<vcount;i++)
s+=polygon[i].y*(polygon[(i-1)].x-polygon[(i+1)%vcount].x);
return s/2;
}
// 如果输入顶点按逆时针排列，返回true
bool isconterclock(int vcount,POINT polygon[])
{
return area_of_polygon(vcount,polygon)>0;
}
void checkconvex(int vcount,POINT polygon[],bool bc[])
{
int i,index=0;
POINT tp=polygon[0];
for(i=1;i<vcount;i++) // 寻找第一个凸顶点
{
if(polygon[i].y<tp.y||(fabs(polygon[i].y- tp.y)<eps&&polygon[i].x<tp.x))
{
tp=polygon[i];
index=i;
}
}
int count=vcount-1;
bc[index]=1;
while(count) // 判断凸凹性
{
if(multiply(polygon[(index+1)%vcount],polygon[(index+2)%vcount],polygon[index%vcount])>=0 )
bc[(index+1)%vcount]=1;
else
bc[(index+1)%vcount]=0;
index++;
count--;
}
}
bool isconvex(int vcount,POINT polygon[])
{
bool bc[1005];
checkconvex(vcount,polygon,bc);
for(int i=0;i<vcount;i++) // 逐一检查顶点，是否全部是凸顶点
if(!bc[i])
return false;
return true;
}
/* 求点p到线段l的最短距离,并返回线段上距该点最近的点np
注意：np是线段l上到点p最近的点，不一定是垂足 */
double dist(POINT p1,POINT p2) // 返回两点之间欧氏距离
{
return( sqrt( (p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y) ) );
}
double dotmultiply(POINT p1,POINT p2,POINT p0)
{
return ((p1.x-p0.x)*(p2.x-p0.x)+(p1.y-p0.y)*(p2.y-p0.y));
}
double relation(POINT p,LINESEG l)
{
LINESEG tl;
tl.s=l.s;
tl.e=p;
return dotmultiply(tl.e,l.e,l.s)/(dist(l.s,l.e)*dist(l.s,l.e));
}
// 求点C到线段AB所在直线的垂足 P
POINT perpendicular(POINT p,LINESEG l)
{
double r=relation(p,l);
POINT tp;
tp.x=l.s.x+r*(l.e.x-l.s.x);
tp.y=l.s.y+r*(l.e.y-l.s.y);
return tp;
}
double ptolinesegdist(POINT p,LINESEG l,POINT &np)
{
double r=relation(p,l);
if(r<0)
{
np=l.s;
return dist(p,l.s);
}
if(r>1)
{
np=l.e;
return dist(p,l.e);
}
np=perpendicular(p,l);
return dist(p,np);
}
/* 计算点到折线集的最近距离,并返回最近点.
注意：调用的是ptolineseg()函数 */
double ptopointset(int vcount,POINT pointset[],POINT p,POINT &q)
{
int i;
double cd=double(2147483647),td;
LINESEG l;
POINT tq,cq;
for(i=0;i<vcount-1;i++)
{
l.s=pointset[i];
l.e=pointset[i+1];
td=ptolinesegdist(p,l,tq);
if(td<cd)
{
cd=td;
cq=tq;
}
}
q=cq;
return cd;
}
/* 判断圆是否在多边形内.ptolineseg()函数的应用2 */
bool CircleInsidePolygon(int vcount,POINT center,double radius,POINT polygon[])
{
POINT q;
double d;
q.x=0;
q.y=0;
d=ptopointset(vcount,polygon,center,q);
if(d<radius||fabs(d-radius)<eps)
return false;
else
return true;
} bool online(LINESEG l,POINT p)
{
return( fabs(multiply(l.e,p,l.s))<eps &&( ( (p.x-l.s.x)*(p.x-l.e.x)<=0 )
&&( (p.y-l.s.y)*(p.y-l.e.y)<=0 ) ) );
}
bool intersect(LINESEG u,LINESEG v)
{
return( (max(u.s.x,u.e.x)>=min(v.s.x,v.e.x))&& //排斥实验
(max(v.s.x,v.e.x)>=min(u.s.x,u.e.x))&&
(max(u.s.y,u.e.y)>=min(v.s.y,v.e.y))&&
(max(v.s.y,v.e.y)>=min(u.s.y,u.e.y))&&
(multiply(v.s,u.e,u.s)*multiply(u.e,v.e,u.s)>=0)&& //跨立实验
(multiply(u.s,v.e,v.s)*multiply(v.e,u.e,v.s)>=0));
}
// (线段u和v相交)&&(交点不是双方的端点) 时返回true
bool intersect_A(LINESEG u,LINESEG v)
{
return ((intersect(u,v))&&
(!online(u,v.s))&&
(!online(u,v.e))&&
(!online(v,u.e))&&
(!online(v,u.s)));
}
int insidepolygon(int vcount,POINT Polygon[],POINT q)
{
int c=0,i,n;
LINESEG l1,l2;
bool bintersect_a,bonline1,bonline2,bonline3;
double r1,r2;
l1.s=q;
l1.e=q;
l1.e.x=double(2147483647);
n=vcount;
for (i=0;i<vcount;i++)
{
l2.s=Polygon[i];
l2.e=Polygon[(i+1)%n];
if(online(l2,q))
return 1; // 如果点在边上，返回1
if ( (bintersect_a=intersect_A(l1,l2))|| // 相交且不在端点
( (bonline1=online(l1,Polygon[(i+1)%n]))&& // 第二个端点在射线上
( (!(bonline2=online(l1,Polygon[(i+2)%n])))&& /* 前一个端点和后一个端点在射线两侧 */
((r1=multiply(Polygon[i],Polygon[(i+1)%n],l1.s)*multiply(Polygon[(i+1)%n],Polygon[(i+2)%n],l1.s))>0) ||
(bonline3=online(l1,Polygon[(i+2)%n]))&& /* 下一条边是水平线，前一个端点和后一个端点在射线两侧 */
((r2=multiply(Polygon[i],Polygon[(i+2)%n],l1.s)*multiply(Polygon[(i+2)%n],
Polygon[(i+3)%n],l1.s))>0)
)
)
) c++;
}
if(c%2 == 1)
return 0;
else
return 2;
}
POINT p[1005];
int main()
{
int i,j,k;
int n;
double r,x,y;
while(cin>>n>>r>>x>>y)
{
if (n<3) break;
double xx,yy;
for (i=0;i<n;i++)
{
scanf("%lf%lf",&xx,&yy);
p[i].x=xx;
p[i].y=yy;
}
if (isconterclock(n,p)==false) //先把点集变成逆时针
{
POINT TMP;
for (i=0;i<n/2;i++)
{
TMP=p[i];
p[i]=p[n-i-1];
p[n-i-1]=TMP;
}
}
bool flag=isconvex(n,p); //判断是否多边形
if (flag)
{
POINT q(x,y);
bool ok=CircleInsidePolygon(n,q,r,p); //判断圆是否被包含，这里的判断方法是圆心到各边的距离中的最小值是否大于等于半径
if (insidepolygon(n,p,q)!=0) //所以需要先判断圆心是否被包含
printf("PEG WILL NOT FIT\n");
else
{
if (ok)
printf("PEG WILL FIT\n");
else
printf("PEG WILL NOT FIT\n");
}
}
else
printf("HOLE IS ILL-FORMED\n");
}
return 0;
}