POJ - 1584 A Round Peg in a Ground Hole（判断凸多边形，点到线段距离，点在多边形内）

http://poj.org/problem?id=1584

题意

按照顺时针或逆时针方向输入一个n边形的顶点坐标集，先判断这个n边形是否为凸包。

再给定一个圆形（圆心坐标和半径），判断这个圆是否完全在n边形内部。

分析

1.判断给出了多边形是不是凸多边形。

2.判断圆包含在凸多边形中：一定要保证圆心在凸多边形里面。然后判断圆心到每条线段的距离要大于等于半径。。

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <queue>
#include <map>
#include <vector>
#include <set>
#include <string>
#include <math.h>

using namespace std;

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;
}
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;
    }
    void input()
    {
        scanf("%lf%lf",&x,&y);
    }
};
struct Line
{
    Point s,e;
    Line(){}
    Line(Point _s,Point _e)
    {
        s = _s;e = _e;
    }
};
//*两点间距离
double dist(Point a,Point b)
{
    return sqrt((a-b)*(a-b));
}
//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;
}
//点到线段的距离
//返回点到线段最近的点
Point NearestPointToLineSeg(Point P,Line L)
{
    Point result;
    double t = ((P-L.s)*(L.e-L.s))/((L.e-L.s)*(L.e-L.s));
    if(t >= 0 && t <= 1)
    {
        result.x = L.s.x + (L.e.x - L.s.x)*t;
        result.y = L.s.y + (L.e.y - L.s.y)*t;
    }
    else
    {
        if(dist(P,L.s) < dist(P,L.e))
            result = L.s;
        else result = L.e;
    }
    return result;
}
//*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;
}
//*判断点在凸多边形内
//点形成一个凸包，而且按逆时针排序（如果是顺时针把里面的<0改为>0）
//点的编号:0~n-1
//返回值：
//-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 inConvexPoly(Point a,Point p[],int n)
{
    for(int i = 0;i < n;i++)
    {
        if(sgn((p[i]-a)^(p[(i+1)%n]-a)) < 0)return -1;
        else if(OnSeg(a,Line(p[i],p[(i+1)%n])))return 0;
    }
    return 1;
}
//*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;
}
//*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
//返回值
//-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 inPoly(Point p,Point poly[],int n)
{
    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;
}
Point p[110];
int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    int n;
    double R,x,y;
    while(scanf("%d",&n) == 1)
    {
        if(n < 3)break;
        scanf("%lf%lf%lf",&R,&x,&y);
        for(int i = 0 ;i < n;i++)
            p[i].input();
        if(!isconvex(p,n))
        {
            printf("HOLE IS ILL-FORMED\n");
            continue;
        }
        Point P = Point(x,y);
        if(inPoly(P,p,n) < 0)
        {
            printf("PEG WILL NOT FIT\n");
            continue;
        }
        bool flag = true;
        for(int i = 0;i < n;i++)
        {
            if(sgn(dist(P,NearestPointToLineSeg(P,Line(p[i],p[(i+1)%n]))) - R) < 0 )
            {
                flag = false;
                break;
            }
        }
        if(flag)printf("PEG WILL FIT\n");
        else printf("PEG WILL NOT FIT\n");
    }
    return 0;
}

原文地址：https://www.cnblogs.com/fht-litost/p/9350179.html