A Round Peg in a Ground Hole(凸包应用POJ 1584)

A Round Peg in a Ground Hole

Time Limit: 1000MS Memory Limit: 10000K

Total Submissions: 5684 Accepted: 1827

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 (xi+1, yi+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

题意：给出一个多边形和一个圆，问是否是凸多边形，若是则再问圆是否在凸多边形内部。

1、判断是否是凸多边形

2、判断点是否在多边形内部

3、判断点到各边的距离是否大于等于半径

#include <iostream>
#include <cstdio>
#include <cmath>
#include <stack>
#include <algorithm>
using namespace std;

const int INF = 0x3f3f3f3f;

const double Pi = 3.141592654;

const double eps = 1e-6;

typedef struct node
{
    double x;
    double y;
} Point;
Point *p;
Point peg;
double pegR;
int n;
double dotdet(node a,node b,node c)//计算点积
{
    return (b.x-a.x)*(c.x-a.x)+(b.y-a.y)*(c.y-a.y);
}
double det(double x1,double y1,double x2,double y2)
{
    return x1*y2-x2*y1;
}

double cross(Point a,Point b,Point c)//计算叉积
{
    return det(b.x-a.x,b.y-a.y,c.x-a.x,c.y-a.y);
}

double Dis(Point a,Point b)//计算距离
{
    return sqrt((b.x-a.x)*(b.x-a.x)+(b.y-a.y)*(b.y-a.y));
}

int precision(double s)//精度控制
{
    if(fabs(s)<=eps)
    {
        return 0;
    }
    return s>0?1:-1;
}

bool JudgeConvex()//判断是否是凸包
{
    int  temp=0;
    int ans;
    for(int i=0; i<n; i++)
    {
        ans=precision(cross(p[i],p[i+1],p[i+2]));
        if(!temp)
        {
            temp=ans;
        }
        if(temp*ans<0)//如果不是同一的角度则不是凸包
        {
            return false;
        }
    }
    return true;
}
double CalAngle(node a,node b)//计算角度
{
    return acos((dotdet(peg,a,b))/(Dis(peg,a)*(Dis(peg,b))));
}
bool JudgeCenter()//判断圆心与凸包的关系
{
    double Angle=0;
    int ans;
    for(int i=0; i<n; i++)
    {
        ans=precision(cross(peg,p[i],p[i+1]));//由叉积判断角度的方向.
        if(ans>=0)
        {
            Angle+=CalAngle(p[i],p[i+1]);
        }
        else
        {
            Angle-=CalAngle(p[i],p[i+1]);
        }
    }
    Angle=fabs(Angle);
    if(precision(Angle)==0)//如果环绕角等于0说明圆心在凸包外侧
    {
        return false;
    }
    else if(precision(Angle-Pi)==0)//如果环绕角为180,圆心在凸包的边上(不包括顶点);
    {
        if(precision(pegR)==0)//只有半径为0的时候才能成立
            return true;
    }
    else if(precision(Angle-2*Pi)==0)//如果环绕角为360,圆心在凸包的里面
    {
        return true;
    }
    else
    {
        if(precision(pegR)==0)//如果环绕角为0-360之间的角度则圆心在凸包的顶点所以只有半径为0的时候才符合
        {
            return true;
        }
    }
    return false;
}

bool JudgeRadius()//判断半径是不是符合:算出圆心到各边的距离和半径进行比较,如果所有的距离都小于半径,则半径是符合的.
{
    for(int i=0;i<n;i++)
    {
        int k=precision(fabs(cross(peg,p[i],p[i+1])/Dis(p[i],p[i+1]))-pegR);
        if(k<0)//如果距离小于半径则不符合
        {
            return false;
        }
    }
    return true;
}

int main()
{
    while(scanf("%d",&n)&&n>=3)
    {
        p =new Point[n+10];
        scanf("%lf %lf %lf",&pegR,&peg.x,&peg.y);
        for(int i=1; i<=n; i++)
        {
            scanf("%lf %lf",&p[i].x,&p[i].y);
        }
        p[0]=p[n];
        p[n+1]=p[1];
        if(!JudgeConvex())
        {
            printf("HOLE IS ILL-FORMED\n");
        }
        else
        {
            bool flag1=JudgeCenter();
            bool flag2=JudgeRadius();
            if(flag1&&flag2)
            {
                printf("PEG WILL FIT\n");
            }
            else
            {
                printf("PEG WILL NOT FIT\n");
            }
        }
    }
    return 0;
}
/*
HOLE IS ILL-FORMED
PEG WILL NOT FIT
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
HOLE IS ILL-FORMED
*/
/*
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
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
4 0.1 1 0.5
0 2
1 0
2 2
1 1
1
*/