A Round Peg in a Ground Hole POJ - 1584(凸边形与圆相交）

A Round Peg in a Ground Hole

 POJ - 1584

题目链接：https://vjudge.net/problem/POJ-1584#author=0

题意：要求钉子要钉入孔内，判断能否在指定点钉入

思路：先判断这些点围成的多边形是不是凸多边形，如果是那么判断圆是否在凸边形里，若在那么就输出“PEG WILL FIT“，不在凸边形里就输出“PEG WILL NOT FIT”，如果都不是凸边形那么输出“HOLE IS ILL-FORMED”

//
// Created by HJYL on 2020/1/22.
//
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
#include<cmath>
using namespace std;
const int maxn=250;
const double eps=1e5;
struct Point{
    double x,y;
    Point(double x = 0, double y = 0):x(x),y(y){}
    Point operator - (const Point &B)
    {
        return Point(x-B.x, y-B.y);
    }
}p[maxn];

int dcmp1(double x)
{
    if (fabs(x)<0) return 0;
    else return x<0?-1:1;
}
bool operator == (const Point &a, const Point &b){
    return dcmp1(a.x-b.x) == 0 && dcmp1(a.y-b.y) == 0;
}

double Cross(Point A,Point B)
{
    return A.x*B.y-A.y*B.x;
}

double dot(Point A,Point B)
{
    return A.x*B.x+A.y*B.y;
}

double Len(Point A)
{
    return sqrt(A.x*A.x+A.y*A.y);
}

bool Istu(Point *p,int n)
{
    p[n]=p[0];
    p[n+1]=p[1];
    int cha=dcmp1(Cross(p[1]-p[0],p[2]-p[1]));
    for(int i=1;i<n;i++)
    {
        int cha1=dcmp1(Cross(p[i+1]-p[i],p[i+2]-p[i+1]));
        if(cha*cha1<0)
            return false;
        cha=cha1;
    }
    return true;
}

int Isintu(Point *p,int n,Point pp)
{
    p[n]=p[0];
    int cha=dcmp1(Cross(p[0]-pp,p[1]-pp));
    for(int i=1;i<n;i++)
    {
        int cha1=dcmp1(Cross(p[i]-pp,p[i+1]-pp));
        if(cha1*cha<0)
            return -1;
        else if(cha1*cha==0)
            return 0;
        cha=cha1;
    }
    return 1;
}

double DistanceToSegment(Point P, Point A, Point B)
{
    if(A == B) return Len(P-A);
    Point v1 = B-A;
    Point v2 = P-A;
    Point v3 = P-B;

    if(dcmp1(dot(v1, v2)) < 0) return Len(v2);
    else if(dcmp1(dot(v1, v3)) > 0) return Len(v3);
    else return fabs(Cross(v1, v2))/ Len(v1);
}

struct Circle{
    Point middle;
    double radius;
}c;

bool Iscircleintu(Point *p,int n,Circle c)
{
    int flag=Isintu(p,n,c.middle);
    if(flag==0)
    {
        if(c.radius == 0) return true;
        else return false;
    }
    else if(flag == 1)
    {
        p[n] = p[0];
        for(int i = 0; i < n; i++)
        {
            if(dcmp1(DistanceToSegment(c.middle, p[i], p[i+1])-c.radius) < 0)
            {
                return false;
            }
        }
        return true;
    }
    else return false;
}


int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        if(n<3)
            break;
        scanf("%lf%lf%lf",&c.radius,&c.middle.x,&c.middle.y);
        for(int i=0;i<n;i++)
        {
            scanf("%lf%lf",&p[i].x,&p[i].y);
        }
        bool result=Istu(p,n);
        if(result)
        {
            int result1=Iscircleintu(p,n,c);
            if(result1)
                printf("PEG WILL FIT\n");
            else
                printf("PEG WILL NOT FIT\n");
        } else
            printf("HOLE IS ILL-FORMED\n");
    }
    return 0;
}

标签：return,Point,int,double,形与,1584,凸边,include
来源： https://www.cnblogs.com/Vampire6/p/12229300.html