POJ 1584-A Round Peg in a Ground Hole（计算几何-凸包、点到线段距离）

ゝ一纸荒年。 2022-07-12 03:40 192阅读 0赞

A Round Peg in a Ground Hole

<table> 
 <tbody> 
  <tr> 
   <td><strong>Time Limit:</strong>&nbsp;1000MS</td> 
   <td>&nbsp;</td> 
   <td><strong>Memory Limit:</strong>&nbsp;10000K</td> 
  </tr> 
  <tr> 
   <td><strong>Total Submissions:</strong>&nbsp;6690</td> 
   <td>&nbsp;</td> 
   <td><strong>Accepted:</strong>&nbsp;2145</td> 
  </tr> 
 </tbody> 
</table>

Description

The DIY Furniture company specializes in assemble-it-yourself furniture kits. Typically, the pieces of wood are attached to one another using a wooden peg that fits into pre-cut holes in each piece to be attached. The pegs have a circular cross-section and so are intended to fit inside a round hole.  
A recent factory run of computer desks were flawed when an automatic grinding machine was mis-programmed. The result is an irregularly shaped hole in one piece that, instead of the expected circular shape, is actually an irregular polygon. You need to figure out whether the desks need to be scrapped or if they can be salvaged by filling a part of the hole with a mixture of wood shavings and glue.  
There are two concerns. First, if the hole contains any protrusions (i.e., if there exist any two interior points in the hole that, if connected by a line segment, that segment would cross one or more edges of the hole), then the filled-in-hole would not be structurally sound enough to support the peg under normal stress as the furniture is used. Second, assuming the hole is appropriately shaped, it must be big enough to allow insertion of the peg. Since the hole in this piece of wood must match up with a corresponding hole in other pieces, the precise location where the peg must fit is known.  
Write a program to accept descriptions of pegs and polygonal holes and determine if the hole is ill-formed and, if not, whether the peg will fit at the desired location. Each hole is described as a polygon with vertices (x1, y1), (x2, y2), . . . , (xn, yn). The edges of the polygon are (xi, yi) to (x i+1, y i+1) for i = 1 . . . n − 1 and (xn, yn) to (x1, y1).

Input

Input consists of a series of piece descriptions. Each piece description consists of the following data:  
Line 1 < nVertices > < pegRadius > < pegX > < pegY >  
number of vertices in polygon, n (integer)  
radius of peg (real)  
X and Y position of peg (real)  
n Lines < vertexX > < vertexY >  
On a line for each vertex, listed in order, the X and Y position of vertex The end of input is indicated by a number of polygon vertices less than 3.

Output

For each piece description, print a single line containing the string:  
HOLE IS ILL-FORMED if the hole contains protrusions  
PEG WILL FIT if the hole contains no protrusions and the peg fits in the hole at the indicated position  
PEG WILL NOT FIT if the hole contains no protrusions but the peg will not fit in the hole at the indicated position

Sample Input

5 1.5 1.5 2.0
    1.0 1.0
    2.0 2.0
    1.75 2.0
    1.0 3.0
    0.0 2.0
    5 1.5 1.5 2.0
    1.0 1.0
    2.0 2.0
    1.75 2.5
    1.0 3.0
    0.0 2.0
    1

Sample Output

HOLE IS ILL-FORMED
    PEG WILL NOT FIT

Source

[Mid-Atlantic 2003][]

## 题目意思： ##

给桌子打孔，孔是不规则的多边形，以坐标形式按顺时针给出这个多边形N个顶点的集合；

圆形的钉子要插入孔中，给出钉子的圆心坐标及其半径。

如果孔是凸多边形，判断钉子是否完全在孔内，若是输出PEG WILL FIT；否则输出PEG WILL NOT FIT；

否则输出HOLE IS ILL-FORMED。

## 解题思路： ##

乱七八糟的套了一把计算几何的模板……大量测试数据在代码后面……

先根据点集判断是否是凸包，然后枚举钉子圆心到凸包各条边的距离，如果距离都小于半径，则钉子能插入孔。

#include<iostream>
    #include<cstdio>
    #include<iomanip>
    #include<cmath>
    #include<cstring>
    #include<algorithm>
    using namespace std;
    const int INF=1e9;
    const int MAXN=200;
    const double eps = 1e-8;
    const double PI = acos(-1.0);
    int sgn(double x)
    {
        if(fabs(x) < eps)return 0;
        if(x < 0)return -1;
        else return 1;
    }
    int n;
    struct Point
    {
        double x,y;
        Point() {}  Point(double _x,double _y)
        {
            x = _x;
            y = _y;
        }
        Point operator -(const Point &b)const
        {
            return Point(x - b.x,y - b.y);//叉积
        }  
        double operator ^(const Point &b)const
        {
            return x*b.y - y*b.x;     //点积
        }
        double operator *(const Point &b)const
        {
            return x*b.x + y*b.y;     //绕原点旋转角度B（弧度值），后x,y的变化
        }
        void transXY(double B)
        {
            double tx = x,ty = y;
            x = tx*cos(B) - ty*sin(B);
            y = tx*sin(B) + ty*cos(B);
        }
    };
    struct Line
    {
        Point s,e;
        Line() {}  Line(Point _s,Point _e)
        {
            s = _s;
            e = _e;
        }
        //两直线相交求交点
        //第一个值为0表示直线重合，为1表示平行，为0表示相交,为2是相交
        //只有第一个值为2时，交点才有意义
        pair<int,Point> operator &(const Line &b)const
        {
            Point res = s;
            if(sgn((s-e)^(b.s-b.e)) == 0)
            {
                if(sgn((s-b.e)^(b.s-b.e)) == 0)
                    return make_pair(0,res);//重合
                else return make_pair(1,res);//平行
            }
            double t = ((s-b.s)^(b.s-b.e))/((s-e)^(b.s-b.e));
            res.x += (e.x-s.x)*t;
            res.y += (e.y-s.y)*t;
            return make_pair(2,res);
        }
    };
    Point list[MAXN];
    //判断凸多边形，允许共线边，点可以是顺时针给出也可以是逆时针给出，点的编号1~n-1
    bool isconvex(Point poly[],int n)
    {
        bool s[3];
        memset(s,false,sizeof(s));
        for(int i = 0; i < n; i++)
        {
            s[sgn( (poly[(i+1)%n]-poly[i])^(poly[(i+2)%n]-poly[i]) )+1] = true;
            if(s[0] && s[2])return false;
        }
        return true;
    }
    double Dot(Point A,Point B)//点积
    {
        return A.x*B.x+A.y*B.y;
    }
    double Length(Point A)
    {
        return sqrt(Dot(A,A));
    }
    double Cross(Point A,Point B)
    {
        return A.x*B.y-A.y*B.x;
    }
    
    double DistanceTosgnment(Point P,Point A,Point B)//点到直线距离
    {
        if(A.x==B.x&&A.y==B.y)
            return Length(P-A);
        if(sgn(Dot(B-A,P-A))<0)return Length(P-A);
        else if(sgn(Dot(B-A,P-B))>0)return Length(P-B);
        else return fabs(Cross(B-A,P-A))/Length(B-A);
    }
    //*判断点在线段上
    bool OnSeg(Point P,Line L)
    {
        return     sgn((L.s-P)^(L.e-P)) == 0 &&
                   sgn((P.x - L.s.x) * (P.x - L.e.x)) <= 0 &&
                   sgn((P.y - L.s.y) * (P.y - L.e.y)) <= 0;
    }
    //*判断线段相交
    bool inter(Line l1,Line l2)
    {
        return max(l1.s.x,l1.e.x) >= min(l2.s.x,l2.e.x) &&
               max(l2.s.x,l2.e.x) >= min(l1.s.x,l1.e.x) &&
               max(l1.s.y,l1.e.y) >= min(l2.s.y,l2.e.y) &&
               max(l2.s.y,l2.e.y) >= min(l1.s.y,l1.e.y) &&
               sgn((l2.s-l1.e)^(l1.s-l1.e))*sgn((l2.e-l1.e)^(l1.s-l1.e)) <= 0 &&
               sgn((l1.s-l2.e)^(l2.s-l2.e))*sgn((l1.e-l2.e)^(l2.s-l2.e)) <= 0;
    }
    int isPointInPolygon(Point p,Point poly[])//点是否在多边形内部
    //*判断点在任意多边形内。射线法，poly[]的顶点数要大于等于3,点的编号0~n-1
    //-1:点在凸多边形外
    //0:点在凸多边形边界上
    //1:点在凸多边形内
    {
    
        int cnt;
        Line ray,side;
        cnt = 0;
        ray.s = p;
        ray.e.y = p.y;
        ray.e.x = -100000000000.0;//-INF,注意取值防止越界
        for(int i = 0; i < n; i++)
        {
            side.s = poly[i];
            side.e = poly[(i+1)%n];
            if(OnSeg(p,side))return 0;
            //如果平行轴则不考虑
            if(sgn(side.s.y - side.e.y) == 0)    continue;
            if(OnSeg(side.s,ray))
            {
                if(sgn(side.s.y - side.e.y) > 0)cnt++;
            }
            else if(OnSeg(side.e,ray))
            {
                if(sgn(side.e.y - side.s.y) > 0)cnt++;
            }
            else if(inter(ray,side))    cnt++;
        }
        if(cnt%2== 1)return 1;
        else return -1;
    }
    
    int main()
    {
    #ifdef ONLINE_JUDGE
    #else
        freopen("F:/cb/read.txt","r",stdin);
        //freopen("F:/cb/out.txt","w",stdout);
    #endif
        ios::sync_with_stdio(false);
        cin.tie(0);
        while(cin>>n)
        {
            if(n<3) break;
            double ra;
            Point r;
            cin>>ra>>r.x>>r.y;
            for(int i=0; i<n; ++i)
                cin>>list[i].x>>list[i].y;
            if(isconvex(list,n))//是凸包
            {
                double dis;//点到直线距离
                bool flag=true;
                int t=isPointInPolygon(r,list);
                if(t!=1)//圆心是否在多边形内部
                    flag=false;
                else//在内部
                {
                    for(int i=0; i<n; ++i)
                    {
                        dis=DistanceTosgnment(r,list[i],list[(i+1)%n]);
                        if(dis<ra)//距离大于半径，无法容纳
                        {
                            flag=false;
                            break;
                        }
                    }
                }
                if(flag)cout<<"PEG WILL FIT"<<endl;
                else cout<<"PEG WILL NOT FIT"<<endl;
            }
            else//不是凸包
                cout<<"HOLE IS ILL-FORMED"<<endl;
        }
        return 0;
    }
    /*
    5 1.5 1.5 2.0
    1.0 1.0
    2.0 2.0
    1.75 2.0
    1.0 3.0
    0.0 2.0
    5 1.5 1.5 2.0
    1.0 1.0
    2.0 2.0
    1.75 2.5
    1.0 3.0
    0.0 2.0
    1
    */

测试数据：

3  0.1  0.2 0.0
    -0.5 1.0
    0.5 -1.0
    0.5 1.0
    3  0.25  0.2 0.0
    -0.5 1.0
    0.5 -1.0
    0.5 1.0
    3 0.1 1.6 1.2
    1.0 1.0
    2.0 1.0
    1.0 2.0
    6 0.1 1.6 1.2
    1.0 1.0
    1.5 1.0
    2.0 1.0
    1.2 1.8
    1.0 2.0
    1.0 1.5
    3 0.1 2.0 2.0
    1.0 1.0
    2.0 1.0
    1.0 2.0
    4  1.0  2.0 1.0
    0.0 0.0
    0.0 4.0
    4.0 4.0
    4.0 0.0
    4  1.0  3.5 1.0
    0.0 0.0
    0.0 4.0
    4.0 4.0
    4.0 0.0
    4  0.2  1.5 1.0
    1.0 1.0
    2.0 2.0
    1.0 3.0
    0.0 2.0
    4  0.4  1.5 1.0
    1.0 1.0
    2.0 2.0
    1.0 3.0
    0.0 2.0
    5  0.2  1.5 2.5
    1.0 1.0
    2.0 2.0
    1.75 2.75
    1.0 3.0
    0.0 2.0
    5  0.2  1.5 2.5
    1.0 1.0
    2.0 2.0
    1.75 2.5
    1.0 3.0
    0.0 2.0
    9 0.2 0.5 2.5
    0.0 0.0
    1.0 0.0
    1.0 1.0
    2.0 1.0
    2.0 0.0
    3.0 0.0
    3.0 5.0
    1.5 5.0
    0.0 5.0
    9 0.2 0.5 2.5
    0.0 0.0
    1.0 0.0
    1.0 -1.0
    2.0 -1.0
    2.0 0.0
    3.0 0.0
    3.0 5.0
    1.5 5.0
    0.0 5.0
    7 0.2 0.5 2.5
    0.0 0.0
    1.0 0.0
    2.0 0.0
    3.0 0.0
    3.0 5.0
    1.5 5.0
    0.0 5.0
    1

运行结果：

PEG WILL FIT
    PEG WILL NOT FIT
    PEG WILL FIT
    PEG WILL FIT
    PEG WILL NOT FIT
    PEG WILL FIT
    PEG WILL NOT FIT
    PEG WILL NOT FIT
    PEG WILL NOT FIT
    PEG WILL FIT
    PEG WILL NOT FIT
    HOLE IS ILL-FORMED
    HOLE IS ILL-FORMED
    PEG WILL FIT

[Mid-Atlantic 2003]: http://poj.org/searchproblem?field=source&key=Mid-Atlantic+2003