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题意

PDF

分析

可以看成直线切割多边形，直接维护。

对每个多边形考虑每条边和每个点即可。

时间复杂度？不过\(n,m \leq 20\)这种数据怎么都过了。

代码

注意初始化的时候的那个初始长方形的点必须按逆时针顺序来插入。

我写反了，导致调了一晚上。

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<set>
#include<map>
#include<queue>
#include<stack>
#include<algorithm>
#include<bitset>
#include<cassert>
#include<ctime>
#include<cstring>
#define rg register
#define il inline
#define co const
template<class T>il T read()
{
    rg T data=0;
    rg int w=1;
    rg char ch=getchar();
    while(!isdigit(ch))
    {
        if(ch=='-')
            w=-1;
        ch=getchar();
    }
    while(isdigit(ch))
    {
        data=data*10+ch-'0';
        ch=getchar();
    }
    return data*w;
}
template<class T>T read(T&x)
{
    return x=read<T>();
}
using namespace std;
typedef long long ll;

co double eps=1e-8;

int dcmp(double x)
{
    return fabs(x)<eps?0:(x<0?-1:1);
}

struct Point
{
    double x,y;
    
    Point(double x=0,double y=0)
    :x(x),y(y){};
};
typedef Point Vector;
typedef vector<Point> Polygon;

Vector operator+(co Vector&A,co Vector&B)
{
    return Vector(A.x+B.x,A.y+B.y);
}

Vector operator-(co Vector&A,co Vector&B)
{
    return Vector(A.x-B.x,A.y-B.y);
}

Vector operator*(co Vector&A,double p)
{
    return Vector(A.x*p,A.y*p);
}

double Dot(co Vector&A,co Vector&B)
{
    return A.x*B.x+A.y*B.y;
}

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

double Length2(co Vector&A)
{
    return Dot(A,A);
}

Point LineLineIntersection(co Point&P,co Vector&v,co Point&Q,co Vector&w)
{
    Vector u=P-Q;
    double t=Cross(w,u)/Cross(v,w);
    return P+v*t;
}

bool OnSegment(co Point&p,co Point&a1,co Point&a2)
{
    return dcmp(Cross(a1-p,a2-p))==0&&dcmp(Dot(a1-p,a2-p))<0;
}

double PolygonArea(co Polygon&poly)
{
    double area=0;
    int n=poly.size();
    for(int i=1;i<n-1;++i)
        area+=Cross(poly[i]-poly[0],poly[(i+1)%n]-poly[0]);
    return area/2;
}

Polygon CutPolygon(co Polygon&poly,co Point&A,co Point&B)
{
    Polygon newpoly;
    int n=poly.size();
    for(int i=0;i<n;++i)
    {
        co Point&C=poly[i];
        co Point&D=poly[(i+1)%n];
        if(dcmp(Cross(B-A,C-A))>=0)
            newpoly.push_back(C);
        if(dcmp(Cross(B-A,C-D))!=0)
        {
            Point ip=LineLineIntersection(A,B-A,C,D-C);
            if(OnSegment(ip,C,D))
                newpoly.push_back(ip);
        }
    }
    return newpoly;
}

int PointInPolygon(co Point&p,co Polygon&v)
{
    int wn=0;
    int n=v.size();
    for(int i=0;i<n;++i)
    {
        if(OnSegment(p,v[i],v[(i+1)%n]))
            return -1;
        int k=dcmp(Cross(v[(i+1)%n]-v[i],p-v[i]));
        int d1=dcmp(v[i].y-p.y);
        int d2=dcmp(v[(i+1)%n].y-p.y);
        if(k>0&&d1<=0&&d2>0)
            ++wn;
        if(k<0&&d2<=0&&d1>0)
            --wn;
    }
    if(wn!=0)
        return 1;
    return 0;
}


bool InCircle(co Point&p,co Point&center,double R)
{
    return dcmp(Length2(p-center)-R*R)<0;
}

int LineCircleIntersection(co Point&A,co Point&B,co Point&C,double r,double&t1,double&t2)
{
    double a=B.x-A.x,
           b=A.x-C.x,
           c=B.y-A.y,
           d=A.y-C.y;
    
    double e=a*a+c*c,
           f=2*(a*b+c*d),
           g=b*b+d*d-r*r,
           delta=f*f-4*e*g;
    if(dcmp(delta)<0)
        return 0;
    if(dcmp(delta)==0)
    {
        t1=t2=-f/(2*e);
        return 1;
    }
    t1=(-f-sqrt(delta))/(2*e);
    t2=(-f+sqrt(delta))/(2*e);
    return 2;
}

bool CircleIntersectSegment(co Point&A,co Point&B,co Point&p,double R)
{
    double t1,t2;
    int c=LineCircleIntersection(A,B,p,R,t1,t2);
    if(c<=1)
        return 0;
    if(dcmp(t1)>0&&dcmp(t1-1)<0)
        return 1;
    if(dcmp(t2)>0&&dcmp(t2-1)<0)
        return 1;
    return 0;
}

vector<Polygon> pieces,new_pieces;

void Cut(int x1,int y1,int x2,int y2)
{
    new_pieces.clear();
    for(int i=0;i<pieces.size();++i)
    {
        Polygon left=CutPolygon(pieces[i],Point(x1,y1),Point(x2,y2));
        Polygon right=CutPolygon(pieces[i],Point(x2,y2),Point(x1,y1));
        if(left.size()>=3)
            new_pieces.push_back(left);
        if(right.size()>=3)
            new_pieces.push_back(right);
//      cerr<<"left="<<endl;
//      for(int j=0;j<left.size();++j)
//          cerr<<" "<<left[j].x<<","<<left[j].y;
//      cerr<<endl;
//      cerr<<"right="<<endl;
//      for(int j=0;j<right.size();++j)
//          cerr<<" "<<right[j].x<<","<<right[j].y;
//      cerr<<endl;
    }
    pieces=new_pieces;
}

bool DiscIntersectPolygon(co Polygon&poly,co Point&p,double R)
{
    if(PointInPolygon(p,poly)!=0)
        return 1;
    if(InCircle(poly[0],p,R))
        return 1;
    int n=poly.size();
    for(int i=0;i<n;++i)
    {
        if(CircleIntersectSegment(poly[i],poly[(i+1)%n],p,R))
            return 1;
        if(InCircle((poly[i]+poly[(i+1)%n])*0.5,p,R))
            return 1;
    }
    return 0;
}

void Query(co Point&p,int R)
{
    vector<double>ans;
    for(int i=0;i<pieces.size();++i)
        if(DiscIntersectPolygon(pieces[i],p,R))
            ans.push_back(fabs(PolygonArea(pieces[i])));
    printf("%d",ans.size());
    sort(ans.begin(),ans.end());
    for(int i=0;i<ans.size();++i)
        printf(" %.2lf",ans[i]);
    printf("\n");
}

int main()
{
//  freopen(".in","r",stdin);
//  freopen(".out","w",stdout);
    int n,m,L,W;
    while(read(n)|read(m)|read(L)|read(W))
    {
//      cerr<<"n="<<n<<" m="<<m<<" L="<<L<<" W="<<W<<endl;
        pieces.clear();
        
        Polygon bbox;
        bbox.push_back(Point(0,0)); // edit 1:must follow the order
        bbox.push_back(Point(L,0));
        bbox.push_back(Point(L,W));
        bbox.push_back(Point(0,W));
        pieces.push_back(bbox);
        
        for(int i=0;i<n;++i)
        {
            int x1,y1,x2,y2;
            read(x1),read(y1),read(x2),read(y2);
//          cerr<<"x1="<<x1<<" y1="<<y1<<" x2="<<x2<<" y2="<<y2<<endl;
            Cut(x1,y1,x2,y2);       
//          for(int i=0;i<pieces.size();++i)
//          {
//              cerr<<i<<" poly="<<endl;
//              for(int j=0;j<pieces[i].size();++j)
//                  cerr<<" "<<pieces[i][j].x<<","<<pieces[i][j].y;
//              cerr<<endl;
//          }
        }
        
        for(int i=0;i<m;++i)
        {
            int x,y,R;
            read(x),read(y),read(R);
            Query(Point(x,y),R);
        }
        printf("\n");
    }
    return 0;
}

UVA12296 Pieces and Discs

标签：题意   box   assert   math   poi   顺序   cross   order   template   

原文地址：https://www.cnblogs.com/autoint/p/10182376.html