17.8.19 校内赛 解题报告【求线段交点+凸包+求多边形面积】【判定点是否在多边形内】【二分答案+半平面交】

Problem 1. land

解题报告
显然，这道题就是线段求交点+凸包+求多边形面积
问题就是我们只有直线求交点的模板和判线段相交的模板，没有直接求线段相交的。我是这样解决的：先判断两线段是否相交，然后再把线段存储为直线，求出两者的交点。
代码如下：

#include<cmath>#include<cstdio>#include<cstring>#include<algorithm>#include<vector>#define Point Vector#define NAME "land"using namespace std;const double eps=1e-8;const int N=1000;int sign(double a){return a>eps?1:(a<-eps?-1:0);}struct Vector{    double x,y;    double len(){return sqrt(x*x+y*y);}    double ang(){return atan2(y,x);}    Vector(double a=.0,double b=.0):x(a),y(b){}    Vector operator+(const Vector &s){return Vector(x+s.x,y+s.y);}    Vector operator-(const Vector &s){return Vector(x-s.x,y-s.y);}    Vector operator*(double s){return Vector(x*s,y*s);}    Vector operator/(double s){return Vector(x/s,y/s);}};struct Seg{    Point A,B;    Seg(){}    Seg(Point a,Point b):A(a),B(b){}}Segs[N+5];struct Line{    Point p;    Vector u;    double ang;    Line(){}    Line(Point p,Vector v):p(p),u(v),ang(u.ang()){}    bool operator<(const Line &s){return ang<s.ang;}    bool operator==(const Line &s){return sign(ang-s.ang)==0;}};bool cmp(Point r,Point s){    if(r.x==s.x)return r.y<s.y;    return r.x<s.x;}double dot(const Vector &a,const Vector &b){return a.x*b.x+a.y*b.y;}//向量点积 double cross(const Vector &a,const Vector &b){return a.x*b.y-a.y*b.x;}//向量叉积bool onleft(Point a,Point b,Point p){    return sign(cross(b-a,p-a))>0;}bool onleft(Line l,Point p){    return sign(cross(l.u,p-l.p))>0;}vector<Point>convex(vector<Point>pts)//求凸包 {    int n=(int)pts.size(),m=0;    vector<Point>cvx;    sort(pts.begin(),pts.end(),cmp);    for(int i=0;i<n;i++)    {        while(m>1&&onleft(cvx[m-2],cvx[m-1],pts[i])){cvx.pop_back();m--;}        cvx.push_back(pts[i]);m++;    }    int k=m;    for(int i=n-2;i>=0;i--)    {        while(m>k&&onleft(cvx[m-2],cvx[m-1],pts[i])){cvx.pop_back();m--;}        cvx.push_back(pts[i]);m++;    }    if(n>1){m--;cvx.pop_back();}    return cvx;}double area(vector<Point>poly)//求多边形的面积 {    double rt=0.0;    int n=(int)poly.size();    for(int i=0;i<n;i++)rt+=cross(poly[i],poly[(i+1)%n]);    return fabs(rt/2.0);}Point line_intersect(Point P,Vector u,Point Q,Vector v) {    double t=cross(Q-P,v)/cross(u,v);    return P+u*t;}Point line_intersect(const Line &a,const Line &b)//求出两直线交点 {    return line_intersect(a.p,a.u,b.p,b.u);}bool seg_intersect(Point A,Point B,Point C,Point D)//判定两线段（AB,CD）是否相交 {    double c1=cross(B-A,C-A),c2=cross(B-A,D-A);    double c3=cross(D-C,A-C),c4=cross(D-C,B-C);    return sign(c1)*sign(c2)<0&&sign(c3)*sign(c4)<0;}int n;vector<Point>pts,poly; int main(){    freopen(NAME".in","r",stdin);    freopen(NAME".out","w",stdout);    scanf("%d",&n);    for(int i=1;i<=n;i++)    {        double x1,y1,x2,y2;        scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);        Segs[i]=Seg(Point(x1,y1),Point(x2,y2));    }    for(int i=1;i<=n-1;i++)    for(int j=i+1;j<=n;j++)    {        if(seg_intersect(Segs[i].A,Segs[i].B,Segs[j].A,Segs[j].B))        pts.push_back(line_intersect(Line((Segs[i].A),Segs[i].B-Segs[i].A),Line((Segs[j].A),Segs[j].B-Segs[j].A)));    }    poly=convex(pts);    double ans=area(poly);    printf("%.2lf",ans);    return 0;}

Problem 2. escape

解题报告
直接套模板上：

#include<cmath>#include<cstdio>#include<cstring>#include<algorithm>#include<vector>#define Point Vector#define NAME "escape"using namespace std;const double eps=1e-8;int sign(double a){return a>eps?1:(a<-eps?-1:0);}struct Vector{    double x,y;    double len(){return sqrt(x*x+y*y);}    double ang(){return atan2(y,x);}    Vector(double a=.0,double b=.0):x(a),y(b){}    Vector operator+(const Vector &s){return Vector(x+s.x,y+s.y);}    Vector operator-(const Vector &s){return Vector(x-s.x,y-s.y);}    Vector operator*(double s){return Vector(x*s,y*s);}    Vector operator/(double s){return Vector(x/s,y/s);}};double dot(const Vector &a,const Vector &b){return a.x*b.x+a.y*b.y;}//向量点积 double cross(const Vector &a,const Vector &b){return a.x*b.y-a.y*b.x;}//向量叉积bool point_on_line(Point p,Point A,Point B)//判定点是否在线段上 {    return sign(cross(B-A,p-A))==0&&sign(dot(A-p,B-p))<=0;}bool in_polygon(Point p,vector<Point>poly)//判定点是否在多边形内 {    int n=(int)poly.size();    int counter=0;    for(int i=0;i<n;++i)    {        Point a=poly[i],b=poly[(i+1)%n];        if(point_on_line(p,a,b))return true; // bounded included        int x=sign(cross(p-a,b-a));        int y=sign(a.y-p.y);        int z=sign(b.y-p.y);        if(x>0&&y<=0&&z>0)counter++;        if(x<0&&z<=0&&y>0)counter--;    }    return counter!=0;}int n,q;vector<Point>poly;int main(){    freopen(NAME".in","r",stdin);    freopen(NAME".out","w",stdout);    scanf("%d%d",&n,&q);    for(int i=1;i<=n;i++)    {        double x,y;        scanf("%lf%lf",&x,&y);        poly.push_back(Point(x,y));    }    while(q--)    {        double x,y;        scanf("%lf%lf",&x,&y);        if(in_polygon(Point(x,y),poly))printf("YES\n");        else printf("NO\n");    }    return 0;}

Problem 3. island

解题报告
这是蓝书上的例题，可以看我的这篇博客
代码如下：

#include<cmath>#include<cstdio>#include<cstring>#include<algorithm>#include<vector>#define NAME "island"#define Point Vectorusing namespace std;const double eps=1e-8;const int N=1e5;int sign(double a){return a>eps?1:(a<-eps?-1:0);}struct Vector{    double x,y;    double len(){return sqrt(x*x+y*y);}    double ang(){return atan2(y,x);}    Vector(double a=.0,double b=.0):x(a),y(b){}    Vector operator+(const Vector &s){return Vector(x+s.x,y+s.y);}    Vector operator-(const Vector &s){return Vector(x-s.x,y-s.y);}    Vector operator*(double s){return Vector(x*s,y*s);}    Vector operator/(double s){return Vector(x/s,y/s);}};Vector Normal(Vector a){double L=a.len();return Vector(-a.y/L,a.x/L);}struct Line{    Point p;    Vector u;    double ang;    Line(){}    Line(Point p,Vector v):p(p),u(v),ang(u.ang()){}    bool operator<(const Line &s){return ang<s.ang;}    bool operator==(const Line &s){return sign(ang-s.ang)==0;}    bool operator<(const Line&L)const{return ang<L.ang;}};double dot(const Vector &a,const Vector &b){return a.x*b.x+a.y*b.y;}//向量点积 double cross(const Vector &a,const Vector &b){return a.x*b.y-a.y*b.x;}//向量叉积Point line_intersect(Point P,Vector u,Point Q,Vector v) {    double t=cross(Q-P,v)/cross(u,v);    return P+u*t;}Point line_intersect(const Line &a,const Line &b)//求出两直线交点 {    return line_intersect(a.p,a.u,b.p,b.u);}bool seg_intersect(Point A,Point B,Point C,Point D)//判定两线段（AB,CD）是否相交 {    double c1=cross(B-A,C-A),c2=cross(B-A,D-A);    double c3=cross(D-C,A-C),c4=cross(D-C,B-C);    return sign(c1)*sign(c2)<0&&sign(c3)*sign(c4)<0;}bool cmp1(Line r,Line s){    return r<s;}bool onleft(Point a,Point b,Point p){    return sign(cross(b-a,p-a))>0;}bool onleft(Line l,Point p){    return sign(cross(l.u,p-l.p))>0;}vector<Point>half_plane_intersect(vector<Line>L){    int n=L.size();    sort(L.begin(),L.end());    int first,last;    vector<Point> p(n);    vector<Line> q(n);    vector<Point> ans;    q[first=last=0]=L[0];    for(int i=1;i<n;i++)    {        while(first<last&&!onleft(L[i],p[last-1]))last--;        while(first<last&&!onleft(L[i],p[first]))first++;        q[++last]=L[i];        if(fabs(cross(q[last].u,q[last-1].u))<eps)        {            last--;            if(onleft(q[last],L[i].p))q[last]=L[i];        }        if(first<last)p[last-1]=line_intersect(q[last-1],q[last]);    }    while(first<last&&!onleft(q[first],p[last-1]))last--;    if(last-first<=1)return ans;    p[last]=line_intersect(q[last],q[first]);    for(int i=first;i<=last;i++)ans.push_back(p[i]);    return ans;}int n;int main(){    freopen(NAME".in","r",stdin);    freopen(NAME".out","w",stdout);    scanf("%d",&n);    vector<Point>p,v,normal;    for(int i=1;i<=n;i++)    {           double x,y;        scanf("%lf%lf",&x,&y);        p.push_back(Point(x,y));    }    for(int i=0;i<=n-1;i++)    {        v.push_back(p[(i+1)%n]-p[i]);        normal.push_back(Normal(v[i]));    }    double lf=0,rg=20000;    while(rg-lf>eps)    {        vector<Line>h;        double mid=lf+(rg-lf)/2.0;        for(int i=0;i<=n-1;i++)        h.push_back(Line(p[i]+normal[i]*mid,v[i]));        vector<Point>poly=half_plane_intersect(h);        if(poly.empty())rg=mid;        else lf=mid;    }    printf("%.2lf",lf);    return 0;}