poj3502 恶心题

巨恶心的一个题::>_<::

题意：给出航班航线和大陆，找航线上距离大陆最远的某一点距离大陆边缘的距离

标准算法：二分答案，从大陆边界向外扩展，扩展出来的面积会覆盖航线。找出航线上最后被覆盖的点即可。

poj讨论版上还有人用模拟退火做的orz

自己的YY算法：如图

找出那些航线与大陆边界的交界点，这些交界点把航线又分成了若干线段。从这些线段上找点即可。

这方法是好想可是不好写啊喂>_<

全写完估计800行啊喂>_<

不写了-_-||

一开始还有脑洞地去求大陆的凸包，实际上那就错了= =。直接按顺序构造多边形即可

上半成品：

#include<vector>
#include<list>
#include<map>
#include<set>
#include<deque>
#include<queue>
#include<stack>
#include<bitset>
#include<algorithm>
#include<functional>
#include<numeric>
#include<utility>
#include<iostream>
#include<sstream>
#include<iomanip>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cctype>
#include<string>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<ctime>
#include<climits>
#include<complex>
#define mp make_pair
#define pb push_back
using namespace std;
const double eps=1e-8;//精度
const double pi=acos(-1.0);//π
const double inf=1e20;//无穷大
const int maxp=1111;//最大点数
/*
    判断d是否在精度内等于0
*/

bool cmp(double x,double y)
{    return x<y;    }

int dblcmp(double d)
{
    if (fabs(d)<eps)return 0;
    return d>eps?1:-1;
}
/*
    求x的平方
*/
inline double sqr(double x){return x*x;}
/*
    点/向量
*/
struct point
{
    double x,y;
    int dis;        //inside(1) OR outside(0) OF the continent
    point(){}
    point(double _x,double _y):x(_x),y(_y){};
    //读入一个点
    void input()
    {
        scanf("%lf%lf",&x,&y);
    }
    //输出一个点
    void output()
    {
        printf("%.2f %.2f
",x,y);
    }
    //判断两点是否相等
    bool operator==(point a)const
    {
        return dblcmp(a.x-x)==0&&dblcmp(a.y-y)==0;
    }
    //判断两点大小
    bool operator<(point a)const
    {
        return dblcmp(a.x-x)==0?dblcmp(y-a.y)<0:x<a.x;
    }
    //点到源点的距离/向量的长度
    double len()
    {
        return hypot(x,y);
    }
    //点到源点距离的平方
    double len2()
    {
        return x*x+y*y;
    }
    //两点间的距离
    double distance(point p)
    {
        return hypot(x-p.x,y-p.y);
    }
    //向量加
    point add(point p)
    {
        return point(x+p.x,y+p.y);
    }
    //向量减
    point sub(point p)
    {
        return point(x-p.x,y-p.y);
    }
    //向量乘
    point mul(double b)
    {
        return point(x*b,y*b);
    }
    //向量除
    point div(double b)
    {
        return point(x/b,y/b);
    }
    //点乘
    double dot(point p)
    {
        return x*p.x+y*p.y;
    }
    //叉乘
    double det(point p)
    {
        return x*p.y-y*p.x;
    }
    //XXXXXXX
    double rad(point a,point b)
    {
        point p=*this;
        return fabs(atan2(fabs(a.sub(p).det(b.sub(p))),a.sub(p).dot(b.sub(p))));
    }
    //截取长度r
    point trunc(double r)
    {
        double l=len();
        if (!dblcmp(l))return *this;
        r/=l;
        return point(x*r,y*r);
    }
    //左转90度
    point rotleft()
    {
        return point(-y,x);
    }
    //右转90度
    point rotright()
    {
        return point(y,-x);
    }
    //绕点p逆时针旋转angle角度
    point rotate(point p,double angle)
    {
        point v=this->sub(p);
        double c=cos(angle),s=sin(angle);
        return point(p.x+v.x*c-v.y*s,p.y+v.x*s+v.y*c);
    }
};
/*
    线段/直线
*/
struct line
{
    point a,b;
    line(){}
    line(point _a,point _b)
    {
        a=_a;
        b=_b;
    }
    //判断线段相等
    bool operator==(line v)
    {
        return (a==v.a)&&(b==v.b);
    }
    //点p做倾斜角为angle的射线
    line(point p,double angle)
    {
        a=p;
        if (dblcmp(angle-pi/2)==0)
        {
            b=a.add(point(0,1));
        }
        else
        {
            b=a.add(point(1,tan(angle)));
        }
    }
    //直线一般式ax+by+c=0
    line(double _a,double _b,double _c)
    {
        if (dblcmp(_a)==0)
        {
            a=point(0,-_c/_b);
            b=point(1,-_c/_b);
        }
        else if (dblcmp(_b)==0)
        {
            a=point(-_c/_a,0);
            b=point(-_c/_a,1);
        }
        else
        {
            a=point(0,-_c/_b);
            b=point(1,(-_c-_a)/_b);
        }
    }
    //读入一个线段
    void input()
    {
        a.input();
        b.input();
    }
    //校准线段两点
    void adjust()
    {
        if (b<a)swap(a,b);
    }
    //线段长度
    double length()
    {
        return a.distance(b);
    }
    //直线倾斜角 0<=angle<180
    double angle()
    {
        double k=atan2(b.y-a.y,b.x-a.x);
        if (dblcmp(k)<0)k+=pi;
        if (dblcmp(k-pi)==0)k-=pi;
        return k;
    }
    //点和线段关系
    //1 在逆时针
    //2 在顺时针
    //3 平行
    int relation(point p)
    {
        int c=dblcmp(p.sub(a).det(b.sub(a)));
        if (c<0)return 1;
        if (c>0)return 2;
        return 3;
    }
    //点是否在线段上
    bool pointonseg(point p)
    {
        return dblcmp(p.sub(a).det(b.sub(a)))==0&&dblcmp(p.sub(a).dot(p.sub(b)))<=0;
    }
    //两线是否平行
    bool parallel(line v)
    {
        return dblcmp(b.sub(a).det(v.b.sub(v.a)))==0;
    }
    //线段和线段关系
    //0 不相交
    //1 非规范相交
    //2 规范相交
    int segcrossseg(line v)
    {
        int d1=dblcmp(b.sub(a).det(v.a.sub(a)));
        int d2=dblcmp(b.sub(a).det(v.b.sub(a)));
        int d3=dblcmp(v.b.sub(v.a).det(a.sub(v.a)));
        int d4=dblcmp(v.b.sub(v.a).det(b.sub(v.a)));
        if ((d1^d2)==-2&&(d3^d4)==-2)return 2;
        return (d1==0&&dblcmp(v.a.sub(a).dot(v.a.sub(b)))<=0||
                d2==0&&dblcmp(v.b.sub(a).dot(v.b.sub(b)))<=0||
                d3==0&&dblcmp(a.sub(v.a).dot(a.sub(v.b)))<=0||
                d4==0&&dblcmp(b.sub(v.a).dot(b.sub(v.b)))<=0);
    }
    //线段和直线v关系
    int linecrossseg(line v)//*this seg v line
    {
        int d1=dblcmp(b.sub(a).det(v.a.sub(a)));
        int d2=dblcmp(b.sub(a).det(v.b.sub(a)));
        if ((d1^d2)==-2)return 2;
        return (d1==0||d2==0);
    }
    //直线和直线关系
    //0 平行
    //1 重合
    //2 相交
    int linecrossline(line v)
    {
        if ((*this).parallel(v))
        {
            return v.relation(a)==3;
        }
        return 2;
    }
    //求两线交点
    point crosspoint(line v)
    {
        double a1=v.b.sub(v.a).det(a.sub(v.a));
        double a2=v.b.sub(v.a).det(b.sub(v.a));
        return point((a.x*a2-b.x*a1)/(a2-a1),(a.y*a2-b.y*a1)/(a2-a1));
    }
    //点p到直线的距离
    double dispointtoline(point p)
    {
        return fabs(p.sub(a).det(b.sub(a)))/length();
    }
    //点p到线段的距离
    double dispointtoseg(point p)
    {
        if (dblcmp(p.sub(b).dot(a.sub(b)))<0||dblcmp(p.sub(a).dot(b.sub(a)))<0)
        {
            return min(p.distance(a),p.distance(b));
        }
        return dispointtoline(p);
    }
    //XXXXXXXX
    point lineprog(point p)
    {
        return a.add(b.sub(a).mul(b.sub(a).dot(p.sub(a))/b.sub(a).len2()));
    }
    //点p关于直线的对称点
    point symmetrypoint(point p)
    {
        point q=lineprog(p);
        return point(2*q.x-p.x,2*q.y-p.y);
    }
};

/*
    多边形
*/
struct polygon
{
    int n;//点个数
    point p[maxp];//顶点
    line l[maxp];//边
    //读入一个多边形
    void input()
    {
        for (int i=0;i<n;i++)
        {
            p[i].input();
        }
    }
    //添加一个点
    void add(point q)
    {
        p[n++]=q;
    }
    //取得边
    void getline()
    {
        for (int i=0;i<n;i++)
        {
            l[i]=line(p[i],p[(i+1)%n]);
        }
    }
    struct cmp
    {
        point p;
        cmp(const point &p0){p=p0;}
        bool operator()(const point &aa,const point &bb)
        {
            point a=aa,b=bb;
            int d=dblcmp(a.sub(p).det(b.sub(p)));
            if (d==0)
            {
                return dblcmp(a.distance(p)-b.distance(p))<0;
            }
            return d>0;
        }
    };
    void norm()
    {
        point mi=p[0];
        for (int i=1;i<n;i++)mi=min(mi,p[i]);
        sort(p,p+n,cmp(mi));
    }
    //求凸包存入多边形convex
    void getconvex(polygon &convex)
    {
        int i,j,k;
        sort(p,p+n);
        convex.n=n;
        for (i=0;i<min(n,2);i++)
        {
            convex.p[i]=p[i];
        }
        if (n<=2)return;
        int &top=convex.n;
        top=1;
        for (i=2;i<n;i++)
        {
            while (top&&convex.p[top].sub(p[i]).det(convex.p[top-1].sub(p[i]))<=0)
                top--;
            convex.p[++top]=p[i];
        }
        int temp=top;
        convex.p[++top]=p[n-2];
        for (i=n-3;i>=0;i--)
        {
            while (top!=temp&&convex.p[top].sub(p[i]).det(convex.p[top-1].sub(p[i]))<=0)
                top--;
            convex.p[++top]=p[i];
        }
    }
    //判断是否凸多边形
    bool isconvex()
    {
        bool s[3];
        memset(s,0,sizeof(s));
        int i,j,k;
        for (i=0;i<n;i++)
        {
            j=(i+1)%n;
            k=(j+1)%n;
            s[dblcmp(p[j].sub(p[i]).det(p[k].sub(p[i])))+1]=1;
            if (s[0]&&s[2])return 0;
        }
        return 1;
    }
    //点与多边形关系
    //0 外部
    //1 内部
    //2 边上
    //3 点上
    int relationpoint(point q)
    {
        int i,j;
        for (i=0;i<n;i++)
        {
            if (p[i]==q)return 3;
        }
        getline();
        for (i=0;i<n;i++)
        {
            if (l[i].pointonseg(q))return 2;
        }
        int cnt=0;
        for (i=0;i<n;i++)
        {
            j=(i+1)%n;
            int k=dblcmp(q.sub(p[j]).det(p[i].sub(p[j])));
            int u=dblcmp(p[i].y-q.y);
            int v=dblcmp(p[j].y-q.y);
            if (k>0&&u<0&&v>=0)cnt++;
            if (k<0&&v<0&&u>=0)cnt--;
        }
        return cnt!=0;
    }
    //线段与多边形关系
    //0 无任何交点
    //1 在多边形内长度为正
    //2 相交或与边平行
    int relationline(line u)
    {
        int i,j,k=0;
        getline();
        for (i=0;i<n;i++)
        {
            if (l[i].segcrossseg(u)==2)return 1;
            if (l[i].segcrossseg(u)==1)k=1;
        }
        if (!k)return 0;
        vector<point>vp;
        for (i=0;i<n;i++)
        {
            if (l[i].segcrossseg(u))
            {
                if (l[i].parallel(u))
                {
                    vp.pb(u.a);
                    vp.pb(u.b);
                    vp.pb(l[i].a);
                    vp.pb(l[i].b);
                    continue;
                }
                vp.pb(l[i].crosspoint(u));
            }
        }
        sort(vp.begin(),vp.end());
        int sz=vp.size();
        for (i=0;i<sz-1;i++)
        {
            point mid=vp[i].add(vp[i+1]).div(2);
            if (relationpoint(mid)==1)return 1;
        }
        return 2;
    }
    //直线u切割凸多边形左侧
    //注意直线方向
    void convexcut(line u,polygon &po)
    {
        int i,j,k;
        int &top=po.n;
        top=0;
        for (i=0;i<n;i++)
        {
            int d1=dblcmp(p[i].sub(u.a).det(u.b.sub(u.a)));
            int d2=dblcmp(p[(i+1)%n].sub(u.a).det(u.b.sub(u.a)));
            if (d1>=0)po.p[top++]=p[i];
            if (d1*d2<0)po.p[top++]=u.crosspoint(line(p[i],p[(i+1)%n]));
        }
    }
    //取得周长
    double getcircumference()
    {
        double sum=0;
        int i;
        for (i=0;i<n;i++)
        {
            sum+=p[i].distance(p[(i+1)%n]);
        }
        return sum;
    }
    //取得面积
    double getarea()
    {
        double sum=0;
        int i;
        for (i=0;i<n;i++)
        {
            sum+=p[i].det(p[(i+1)%n]);
        }
        return fabs(sum)/2;
    }
    bool getdir()//1代表逆时针 0代表顺时针
    {
        double sum=0;
        int i;
        for (i=0;i<n;i++)
        {
            sum+=p[i].det(p[(i+1)%n]);
        }
        if (dblcmp(sum)>0)return 1;
        return 0;
    }
    //取得重心
    point getbarycentre()
    {
        point ret(0,0);
        double area=0;
        int i;
        for (i=1;i<n-1;i++)
        {
            double tmp=p[i].sub(p[0]).det(p[i+1].sub(p[0]));
            if (dblcmp(tmp)==0)continue;
            area+=tmp;
            ret.x+=(p[0].x+p[i].x+p[i+1].x)/3*tmp;
            ret.y+=(p[0].y+p[i].y+p[i+1].y)/3*tmp;
        }
        if (dblcmp(area))ret=ret.div(area);
        return ret;
    }
    //点在凸多边形内部的判定
    int pointinpolygon(point q)
    {
        if (getdir())reverse(p,p+n);
        if (dblcmp(q.sub(p[0]).det(p[n-1].sub(p[0])))==0)
        {
            if (line(p[n-1],p[0]).pointonseg(q))return n-1;
            return -1;
        }
        int low=1,high=n-2,mid;
        while (low<=high)
        {
            mid=(low+high)>>1;
            if (dblcmp(q.sub(p[0]).det(p[mid].sub(p[0])))>=0&&dblcmp(q.sub(p[0]).det(p[mid+1].sub(p[0])))<0)
            {
                polygon c;
                c.p[0]=p[mid];
                c.p[1]=p[mid+1];
                c.p[2]=p[0];
                c.n=3;
                if (c.relationpoint(q))return mid;
                return -1;
            }
            if (dblcmp(q.sub(p[0]).det(p[mid].sub(p[0])))>0)
            {
                low=mid+1;
            }
            else
            {
                high=mid-1;
            }
        }
        return -1;
    }
};

struct polygon P[100],R[100];       //P:continent    R:convex of the continent
struct point key[100];              //key point
int T,C,N,M;


int main()
{
    freopen("in.txt","r",stdin);

    cin>>T;
    while (T--)
    {
        cin>>C>>N;
        for (int i=1;i<=N;i++)
            key[i].input();
        for (int i=1;i<=C;i++)
        {
            cin>>M;
            P[i].n=M;
            P[i].input();
            P[i].getline();
            for (int j=1;j<=N;j++)
            {
                if (P[i].relationpoint(key[j])==0)  //outside
                    key[j].dis=0;
                else
                    key[j].dis=1;
            }
        }
        double mmx=0,dist;
        struct line maxLN;
        for (int i=1;i<N;i++)
        {
            struct line LN;
            LN.a=key[i];    LN.b=key[i+1];
            int tn=0;
            double tp[100];
            for (int j=1;j<=C;j++)      //continent j
            {
                for (int k=0;k<P[j].n;k++)  //line k in continent j
                {
                    line TL=P[j].l[k];
                    if (LN.segcrossseg(TL)!=0)
                    {
                        point tm=LN.crosspoint(TL); //crosspoint of continent && flight route
                        tp[tn]=key[i].distance(tm);
                        tn++;
                    }
                }
            }
            for (int ii=0;ii<tn;ii++)  printf("%.6f ",tp[ii]);  printf("
");
            sort(tp,tp+tn,cmp);
            for (int ii=0;ii<tn;ii++)  printf("%.6f ",tp[ii]);  printf("
");
            if (key[i].dis==0)  //outside
            {
                for (int j=0;j<tn;j=j+2)
                {
                    double TM;
                    if (j==0)   TM=tp[j];   else TM=tp[j]-tp[j-1];
                    if (TM>mmx) 
                    {
                        maxLN=LN;
                        mmx=TM;
                        if (j==0) dist=mmx/2; else dist=tp[j-1]+mmx/2;
                    }
                }
            }
            else    //inside
            {
                for (int j=1;j<tn;j=j+2)
                {
                    double TM;
                    TM=tp[j]-tp[j-1];
                    if (TM>mmx) 
                    {
                        maxLN=LN;
                        mmx=TM;
                        dist=tp[j-1]+mmx/2;
                    }
                }
            }
        }
        double dx=maxLN.a.distance(maxLN.b);    dx=dist/dx;
        point A=maxLN.a,B=maxLN.b;
        point Ans=(A.x+(B.x-A.x)*dx,A.y+(B.y-A.y)*dy);
        
                
        printf("%.6f
",mmx/2);
    }
    return 0;
}

果然计算几何就是坑坑坑坑T^T

在 http://poj.org/showcontest?contest_id=1477上面找的，实际上这一套题除了A剩下的都是坑= =