Codeforces problem 67E(多边形求内核的应用)

最新推荐文章于 2019-02-28 19:28:00 发布

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本文介绍了一种计算多边形核的方法，包括区域面积、周长等关键属性的计算。通过使用半平面相交算法来确定多边形核的存在与否，并计算其几何属性。

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题目：E. Save the City!

/*

Goujinping 2013.4.12  NEFU

The masterplate of Polygon kernel.

Now the global variable Area stand for the area of Polygon  kernel

In most case,the problem let us judge whether the Polygon kernel
exist or not and calculate the area,perimeter,or other constants
about the Polygon kernel.

*/

#include <math.h>
#include <stdio.h>
#include <iostream>
#include <algorithm>

using namespace std;

const int N=11111;
const double EPS = 1e-8;

typedef double DIY;

DIY Area,Length;

struct Point
{
    DIY x,y;
    Point() {}
    Point(DIY _x,DIY _y):x(_x),y(_y) {}
} p[N];

Point MakeVector(Point &P,Point &Q)
{
    return Point(Q.x-P.x,Q.y-P.y);
}

DIY CrossProduct(Point P,Point Q)
{
    return P.x*Q.y-P.y*Q.x;
}

DIY MultiCross(Point P,Point Q,Point R)
{
    return CrossProduct(MakeVector(Q,P),MakeVector(Q,R));
}

struct halfPlane
{
    Point s,t;
    DIY angle;
    halfPlane() {}
    halfPlane(Point _s,Point _t):s(_s),t(_t) {}
    halfPlane(DIY sx,DIY sy,DIY tx,DIY ty):s(sx,sy),t(tx,ty) {}
    void GetAngle()
    {
        angle=atan2(t.y-s.y,t.x-s.x);
    }
} hp[N],q[N];

Point IntersectPoint(halfPlane P,halfPlane Q)
{
    DIY a1=CrossProduct(MakeVector(P.s,Q.t),MakeVector(P.s,Q.s));
    DIY a2=CrossProduct(MakeVector(P.t,Q.s),MakeVector(P.t,Q.t));
    return Point((P.s.x*a2+P.t.x*a1)/(a2+a1),(P.s.y*a2+P.t.y*a1)/(a2+a1));
}

bool cmp(halfPlane P,halfPlane Q)
{
    if(fabs(P.angle-Q.angle)<EPS)
        return MultiCross(P.s,P.t,Q.s)>0;
    return P.angle<Q.angle;
}

bool IsParallel(halfPlane P,halfPlane Q)
{
    return fabs(CrossProduct(MakeVector(P.s,P.t),MakeVector(Q.s,Q.t)))<EPS;
}

void HalfPlaneIntersect(int n,int &m)
{
    sort(hp,hp+n,cmp);
    int i,l=0,r=1;
    for(m=i=1; i<n; ++i)
        if(hp[i].angle-hp[i-1].angle>EPS) hp[m++]=hp[i];
    n=m; m=0;
    q[0]=hp[0];q[1]=hp[1];
    for(i=2; i<n; i++)
    {
        if(IsParallel(q[r],q[r-1])||IsParallel(q[l],q[l+1])) return;
        while(l<r&&MultiCross(hp[i].s,hp[i].t,IntersectPoint(q[r],q[r-1]))>0) --r;
        while(l<r&&MultiCross(hp[i].s,hp[i].t,IntersectPoint(q[l],q[l+1]))>0) ++l;
        q[++r]=hp[i];
    }
    while(l<r&&MultiCross(q[l].s,q[l].t,IntersectPoint(q[r],q[r-1]))>0) --r;
    while(l<r&&MultiCross(q[r].s,q[r].t,IntersectPoint(q[l],q[l+1]))>0) ++l;
    q[++r]=q[l];
    for(i=l; i<r; ++i)
        p[m++]=IntersectPoint(q[i],q[i+1]);
}

void Solve(Point *p,int n,int &m)
{
    int i,j;
    Point a,b;
    p[n]=p[0];
    for(i=0;i<n;i++)
    {
        hp[i]=halfPlane(p[(i+1)%n],p[i]);
        hp[i].GetAngle();
    }
    a=p[0],b=p[1];
    HalfPlaneIntersect(n,m);

    Area=0;Length=0;

    if(m>2)
    {
        if(a.x>b.x) swap(a,b);
        int f2=0,f1=0;
        for(int i=0; i<m; i++)
        {
            if(p[i].y==a.y&&p[(i+1)%m].y==a.y)
            {
                f2=1;
                a=p[i];
                b=p[(i+1)%m];
                break;
            }
            else if(p[i].y==a.y)
                f1=1;
        }
        if(f1&&!f2)
            Length=1;
        else if(f2)
        {
            if(a.x>b.x)
                swap(a,b);
            Length=(int)(b.x-a.x)+1;
        }
    }

    if(m>2)
    {
        p[m]=p[0];
        for(i=0;i<m;++i)
            Area+=CrossProduct(p[i],p[i+1]);
        if(Area<0) Area=-Area;
    }
    Area/=2.0;
}

int main()
{
    int n,m;
    while(cin>>n)
    {
        for(int i=0; i<n; i++)
            cin>>p[i].x>>p[i].y;
        Solve(p,n,m);
        cout<<Length<<endl;
    }
    return 0;
}