LA 2797

题目链接

题意：训练指南283页；

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>
#include <map>
#include <algorithm>
#include <set>
#define MM(a) memset(a,0,sizeof(a))
typedef long long ll;
typedef unsigned long long ULL;
const double eps = 1e-14;
const int inf = 0x3f3f3f3f;
const double pi=acos(-1);
using namespace std;

struct Point {
    double x, y;
    double ang;
    Point() {}
    Point(double x,double y) {
        this->x = x;
        this->y = y;
    }
    void read() {
        scanf("%lf %lf", &x, &y);
    }
    bool operator <(const Point w) const
    {
        if(this->x==w.x)  return this->y<w.y;
        else   return this->x<w.x;
    }
};
typedef Point Vector;
/*
struct Line{
    Point a;
    Point v,nor;
    double ang;
    Line(){};
    Line(Point u,Vector w):a(u),v(w){};
    bool operator <(const Line q) const{
       return this->ang<q.ang;
    }
};

struct Circle {
    Point c;
    double r;
    Circle(){};
    Circle(Point c, double r) {
        this->c = c;
        this->r = r;
    }
    Point point(double a) {
        return Point(c.x + cos(a) * r, c.y + sin(a) * r);
    }

};
*/
Vector operator + (Vector A, Vector B) {
    return Vector(A.x + B.x, A.y + B.y);
}

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

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

Vector operator / (Vector A, double p) {
    return Vector(A.x / p, A.y / p);
}

const double PI = acos(-1.0);

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

bool operator == (const Point& a, const Point& b) {
    return dcmp(a.x - b.x) == 0 && dcmp(a.y - b.y) == 0;
}

bool operator < (const Point& a, const Point& b) {
    return a.x < b.x || (a.x == b.x && a.y < b.y);
}

double torad(double ang)
{
    return ang/180*pi;
}

double Dot(Vector A, Vector B) {return A.x * B.x + A.y * B.y;} //点积
double Length(Vector A) {return sqrt(Dot(A, A));} //向量的模
double Angle(Vector A, Vector B) {return acos(Dot(A, B) / Length(A) / Length(B));} //向量夹角
double Cross(Vector A, Vector B) {return A.x * B.y - A.y * B.x;} //叉积
double Area2(Point A, Point B, Point C) {return Cross(B - A, C - A);} //有向面积
double angle(Vector v) {return atan2(v.y, v.x);}

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

Vector Rotate(Vector A, double rad) {
    return Vector(A.x * cos(rad) - A.y * sin(rad), A.x * sin(rad) + A.y * cos(rad));
}

double DistanceToLine(Point P, Point A, Point B) {
    Vector v1 = B - A, v2 = P - A;
    return fabs(Cross(v1, v2)) / Length(v1);
}

//线段的规范相交
bool SegmentProperIntersection(Point a1, Point a2, Point b1, Point b2) {
    double c1 = Cross(a2 - a1, b1 - a1), c2 = Cross(a2 - a1, b2 - a1),
            c3 = Cross(b2 - b1, a1 - b1), c4 = Cross(b2 - b1, a2 - b1);
    return dcmp(c1) * dcmp(c2) < 0 && dcmp(c3) * dcmp(c4) < 0;
}

//点在线段上(不含端点)
bool OnSegment(Point p, Point a1, Point a2) {
    return dcmp(Cross(a1 - p, a2 - p)) == 0 && dcmp(Dot(a1 - p, a2 - p))<0;
}

//线段不规范相交 （自己写的）
bool SegmentinProperIntersection(Point a1, Point a2, Point b1, Point b2)
{
    if(SegmentProperIntersection(a1, a2, b1,b2))
       return 1;
    if(OnSegment(b1, a1, a2))
       return 1;
    if(OnSegment(b2, a1, a2))
       return 1;
    return 0;
}

Point p[1005];
Vector v[105];
int n,num,f[505][505],vis[505];
vector<Point> q;

bool Onanysegment(Point w)
{
   for(int i=0;i<n;i++)
        if(OnSegment(w,p[i],p[i+n]))
            return true;
   return false;
}

bool Intercetwithangsegment(Point a,Point b)
{
    for(int i=0;i<n;i++)
        if(SegmentProperIntersection(a,b,p[i],p[i+n]))
             return true;
    return false;
}

void init()
{
   memset(f,0,sizeof(f));
   memset(vis,0,sizeof(vis));
   q.clear();

   p[2*n]=Point(0,0);p[2*n+1]=Point(1000,1000);
   q.push_back(p[2*n]);q.push_back(p[2*n+1]);

   for(int i=0;i<=2*n-1;i++)
     if(!Onanysegment(p[i]))
       q.push_back(p[i]);//对于处在线段中间的点，不进入构图，否则会错，想像一下，两条线段共端点且共线

   num=q.size();
   for(int i=0;i<num;i++)
     for(int j=i+1;j<num;j++)
          if(!Intercetwithangsegment(q[i],q[j]))
            f[j][i]=f[i][j]=1;//说明这两个点可以直接到达

   //for(int i=0;i<num;i++)
     // for(int j=0;j<num;j++)
       //cout<<i<<" "<<j<<" "<<f[i][j]<<endl;
}

bool dfs(int cur)
{
   // printf("cur:%d
",cur);
    if(cur==1) return true;
    vis[cur]=1;
    for(int i=0;i<num;i++)
       if(!vis[i]&&f[cur][i]&&dfs(i))//dfs的标准格式
          return true;
    return false;
}

void solve()
{
    if(dfs(0)) printf("no
");
    else printf("yes
");
}

int main()
{
    while(~scanf("%d",&n)&&n)
    {
       for(int i=0;i<n;i++)
            {
                p[i].read();p[i+n].read();
                v[i]=(p[i+n]-p[i])/(Length(p[i+n]-p[i]));
                p[i]=p[i]-v[i]*1e-5;
                p[i+n]=p[i+n]+v[i]*1e-5;//进行端点的微小扰动
            }
       init();
       solve();
    }
    return 0;
}

wa代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>
#include <map>
#include <algorithm>
#include <set>
#define MM(a) memset(a,0,sizeof(a))
typedef long long ll;
typedef unsigned long long ULL;
const double eps = 1e-14;
const int inf = 0x3f3f3f3f;
const double pi=acos(-1);
using namespace std;

struct Point {
    int x, y;
    double ang;
    Point() {}
    Point(int x,int y) {
        this->x = x;
        this->y = y;
    }
    void read() {
        scanf("%lf%lf", &x, &y);
    }
    bool operator <(const Point w) const
    {
        if(this->x==w.x)  return this->y<w.y;
        else   return this->x<w.x;
    }
};
typedef Point Vector;
/*
struct Line{
    Point a;
    Point v,nor;
    double ang;
    Line(){};
    Line(Point u,Vector w):a(u),v(w){};
    bool operator <(const Line q) const{
       return this->ang<q.ang;
    }
};

struct Circle {
    Point c;
    double r;
    Circle(){};
    Circle(Point c, double r) {
        this->c = c;
        this->r = r;
    }
    Point point(double a) {
        return Point(c.x + cos(a) * r, c.y + sin(a) * r);
    }

};
*/
Vector operator + (Vector A, Vector B) {
    return Vector(A.x + B.x, A.y + B.y);
}

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

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

Vector operator / (Vector A, double p) {
    return Vector(A.x / p, A.y / p);
}

const double PI = acos(-1.0);

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

bool operator == (const Point& a, const Point& b) {
    return dcmp(a.x - b.x) == 0 && dcmp(a.y - b.y) == 0;
}

bool operator < (const Point& a, const Point& b) {
    return a.x < b.x || (a.x == b.x && a.y < b.y);
}

double torad(double ang)
{
    return ang/180*pi;
}

double Dot(Vector A, Vector B) {return A.x * B.x + A.y * B.y;} //点积
double Length(Vector A) {return sqrt(Dot(A, A));} //向量的模
double Angle(Vector A, Vector B) {return acos(Dot(A, B) / Length(A) / Length(B));} //向量夹角
double Cross(Vector A, Vector B) {return A.x * B.y - A.y * B.x;} //叉积
double Area2(Point A, Point B, Point C) {return Cross(B - A, C - A);} //有向面积
double angle(Vector v) {return atan2(v.y, v.x);}

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

Vector Rotate(Vector A, double rad) {
    return Vector(A.x * cos(rad) - A.y * sin(rad), A.x * sin(rad) + A.y * cos(rad));
}

double DistanceToLine(Point P, Point A, Point B) {
    Vector v1 = B - A, v2 = P - A;
    return fabs(Cross(v1, v2)) / Length(v1);
}

//线段的规范相交
bool SegmentProperIntersection(Point a1, Point a2, Point b1, Point b2) {
    int c1 = Cross(a2 - a1, b1 - a1), c2 = Cross(a2 - a1, b2 - a1),
            c3 = Cross(b2 - b1, a1 - b1), c4 = Cross(b2 - b1, a2 - b1);
    return dcmp(c1) * dcmp(c2) < 0 && dcmp(c3) * dcmp(c4) < 0;
}

//点在线段上(含端点)
bool OnSegment(Point p, Point a1, Point a2) {
    return dcmp(Cross(a1 - p, a2 - p)) == 0 && dcmp(Dot(a1 - p, a2 - p))<=0;
}

//线段不规范相交 （自己写的）
bool SegmentinProperIntersection(Point a1, Point a2, Point b1, Point b2)
{
    if(SegmentProperIntersection(a1, a2, b1,b2))
       return 1;
    if(OnSegment(b1, a1, a2))
       return 1;
    if(OnSegment(b2, a1, a2))
       return 1;
    return 0;
}
/*
Vector AngleBisector(Point p, Vector v1, Vector v2){//给定两个向量，求角平分线
    double rad = Angle(v1, v2);
    return Rotate(v1, dcmp(Cross(v1, v2)) * 0.5 * rad);
}

//求线与x轴的真实角(0<=X<180)
double RealAngleWithX(Vector a){
    Vector b(1, 0);
    if (dcmp(Cross(a, b)) == 0) return 0.0;
    else if (dcmp(Dot(a, b) == 0)) return 90.0;
    double rad = Angle(a, b);
    rad = (rad / PI) * 180.0;
    if (dcmp(a.y) < 0) rad = 180.0 - rad;
    return rad;
}

//求直线与圆的交点
int getLineCircleIntersection(Point p, Vector v, Circle c, vector<Point> &sol) {
    double a1 = v.x, b1 = p.x - c.c.x, c1 = v.y, d1 = p.y - c.c.y;
    double e1 = a1 * a1 +  c1 * c1, f1 = 2 * (a1 * b1 + c1 * d1), g1 = b1 * b1 + d1 * d1 - c.r * c.r;
    double delta = f1 * f1 - 4 * e1 * g1, t;
    if(dcmp(delta) < 0) return 0;
    else if(dcmp(delta) == 0){
        t = (-f1) / (2 * e1);
        sol.push_back(p + v * t);
        return 1;
    } else{
        t = (-f1 + sqrt(delta)) / (2 * e1); sol.push_back(p + v * t);
        t = (-f1 - sqrt(delta)) / (2 * e1); sol.push_back(p + v * t);
        return 2;
    }
}

//两圆相交
int getCircleCircleIntersection(Circle C1, Circle C2, vector<Point> &sol) {
    double d = Length(C1.c - C2.c);
    if (dcmp(d) == 0) {
        if (dcmp(C1.r - C2.r) == 0) return -1; // 重合
        return 0;
    }
    if (dcmp(C1.r + C2.r - d) < 0) return 0;
    if (dcmp(fabs(C1.r - C2.r) - d) > 0) return 0;
    double a = angle(C2.c - C1.c);
    double da = acos((C1.r * C1.r + d * d - C2.r * C2.r) / (2 * C1.r * d));
    Point p1 = C1.point(a - da), p2 = C1.point(a + da);
    sol.push_back(p1);
    if(p1 == p2) return 1;
    sol.push_back(p2);
    return 2;

}

//点到圆的切线
int getTangents(Point p, Circle C, Vector *v) {
    Vector u = C.c - p;
    double dist = Length(u);
    if (dist < C.r) return 0;
    else if (dcmp(dist - C.r) == 0) {
        v[0] = Rotate(u, PI / 2);
        return 1;
    } else {
        double ang = asin(C.r / dist);
        v[0] = Rotate(u, -ang);
        v[1] = Rotate(u, +ang);
        return 2;
    }
}

//两圆公切线
//a[i], b[i]分别是第i条切线在圆A和圆B上的切点
int getCircleTangents(Circle A, Circle B, Point *a, Point *b) {
    int cnt = 0;
    if (A.r < B.r) { swap(A, B); swap(a, b); }
    //圆心距的平方
    double d2 = (A.c.x - B.c.x) * (A.c.x - B.c.x) + (A.c.y - B.c.y) * (A.c.y - B.c.y);
    double rdiff = A.r - B.r;
    double rsum = A.r + B.r;
    double base = angle(B.c - A.c);
    //重合有无限多条
    if (d2 == 0 && dcmp(A.r - B.r) == 0) return -1;
    //内切
    if (dcmp(d2 - rdiff * rdiff) == 0) {
        a[cnt] = A.point(base);
        b[cnt] = B.point(base);
        cnt++;
        return 1;
    }
    //有外公切线
    double ang = acos((A.r - B.r) / sqrt(d2));
    a[cnt] = A.point(base + ang); b[cnt] = B.point(base + ang); cnt++;
    a[cnt] = A.point(base - ang); b[cnt] = B.point(base - ang); cnt++;

    //一条内切线，两条内切线
    if (dcmp(d2 - rsum*rsum) == 0) {
        a[cnt] = A.point(base); b[cnt] = B.point(PI + base); cnt++;
    } else if (dcmp(d2 - rsum*rsum) > 0) {
        double ang = acos((A.r + B.r) / sqrt(d2));
        a[cnt] = A.point(base + ang); b[cnt] = B.point(base + ang); cnt++;
        a[cnt] = A.point(base - ang); b[cnt] = B.point(base - ang); cnt++;
    }
    return cnt;
}

//三角形外切圆
Circle CircumscribedCircle(Point p1, Point p2, Point p3) {
    double Bx = p2.x - p1.x, By = p2.y - p1.y;
    double Cx = p3.x - p1.x, Cy = p3.y - p1.y;
    double D = 2 * (Bx * Cy - By * Cx);
    double cx = (Cy * (Bx * Bx + By * By) - By * (Cx * Cx + Cy * Cy)) / D + p1.x;
    double cy = (Bx * (Cx * Cx + Cy * Cy) - Cx * (Bx * Bx + By * By)) / D + p1.y;
    Point p = Point(cx, cy);
    return Circle(p, Length(p1 - p));
}

//三角形内切圆
Circle InscribedCircle(Point p1, Point p2, Point p3) {
    double a = Length(p2 - p3);
    double b = Length(p3 - p1);
    double c = Length(p1 - p2);
    Point p = (p1 * a + p2 * b + p3 * c) / (a + b + c);
    return Circle(p, DistanceToLine(p, p1, p2));
}

//求经过点p1,与直线(p2, w)相切,半径为r的一组圆
int CircleThroughAPointAndTangentToALineWithRadius(Point p1, Point p2, Vector w, double r, vector<Point> &sol) {
    Circle c1 = Circle(p1, r);
    double t = r / Length(w);
    Vector u = Vector(-w.y, w.x);
    Point p4 = p2 + u * t;
    int tot = getLineCircleIntersection(p4, w, c1, sol);
    u = Vector(w.y, -w.x);
    p4 = p2 + u * t;
    tot += getLineCircleIntersection(p4, w, c1, sol);
    return tot;
}

//给定两个向量，求两向量方向内夹着的圆的圆心。圆与两线均相切，圆的半径已给定
Point Centre_CircleTangentTwoNonParallelLineWithRadius(Point p1, Vector v1, Point p2, Vector v2, double r){
    Point p0 = GetLineIntersection(p1, v1, p2, v2);
    Vector u = AngleBisector(p0, v1, v2);
    double rad = 0.5 * Angle(v1, v2);
    double l = r / sin(rad);
    double t = l / Length(u);
    return p0 + u * t;
}

//求与两条不平行的直线都相切的4个圆，圆的半径已给定
int CircleThroughAPointAndTangentALineWithRadius(Point p1, Vector v1, Point p2, Vector v2, double r, Point *sol) {
    int ans = 0;
    sol[ans++] = Centre_CircleTangentTwoNonParallelLineWithRadius(p1, v1, p2, v2, r);
    sol[ans++] = Centre_CircleTangentTwoNonParallelLineWithRadius(p1, v1 * -1, p2, v2, r);
    sol[ans++] = Centre_CircleTangentTwoNonParallelLineWithRadius(p1, v1, p2, v2 * -1, r);
    sol[ans++] = Centre_CircleTangentTwoNonParallelLineWithRadius(p1, v1 * -1, p2, v2 * -1, r);
    return ans;
}

//求与两个相离的圆均外切的一组圆，三种情况
int CircleTangentToTwoDisjointCirclesWithRadius(Circle c1, Circle c2, double r, Point *sol){
    double dis1 = c1.r + r + r + c2.r;
    double dis2= Length(c1.c - c2.c);
    if(dcmp(dis1 - dis2) < 0) return 0;
    Vector u = c2.c - c1.c;
    double t = (r + c1.r) / Length(u);
    if(dcmp(dis1 - dis2)==0){
        Point p0 = c1.c + u * t;
        sol[0] = p0;
        return 1;
    }
    double aa = Length(c1.c - c2.c);
    double bb = r + c1.r, cc = r + c2.r;
    double rad = acos((aa * aa + bb * bb - cc * cc) / (2 * aa * bb));
    Vector w = Rotate(u, rad);
    Point p0 = c1.c + w * t;
    sol[0] = p0;
    w = Rotate(u, -rad);
    p0 = c1.c + w * t;
    sol[1] = p0;
    return 2;
}
//判断点与圆的位置关系（自己写的）
int  pointincircle(Point a,Circle o)
{
     double l=Length(o.c-a);
     if(dcmp(l-o.r)>0)
           return 1;
     else if(dcmp(l-o.r)==0)
           return 0;
     else if(dcmp(l-o.r)<0)
           return -1;
}
int ConvexHull(Point *p, int n, Point* ch)         //求凸包
 {
     sort(p, p + n);//先按照x，再按照y
     int m = 0;
     for(int i = 0; i < n; i++)
     {
         while(m > 1 && Cross(ch[m-1] - ch[m-2], p[i] - ch[m-2]) <= 0) m--;
        ch[m++] = p[i];
    }
    int k = m;
     for(int i = n-2; i >= 0; i--)
    {
          while(m > k && Cross(ch[m-1] - ch[m-2], p[i] - ch[m-2]) <= 0) m--;
          ch[m++] = p[i];
      }
      if(n > 1) m--;
      return m;
}

double Polygonarea(Point *p,int n)
{
   double area=0;
   for(int i=1;i<n-1;i++)
        area+=Cross(p[i]-p[0],p[i+1]-p[0]);
   return area/2;
}

//判断点是否在凸多边形内，注意是凸多变形，不是多边形
int Pointinpolygon(Point p,Point *q,int m)
{
    for(int i=0;i<m;i++)
       if(Cross(q[i+1]-q[i],p-q[i])<=0)
          return 0;
    return 1;
}

Point ne[105],tubao[105];

void rotating_calipers(Point *p,int k)
{
    double ans=0;
    p[k]=p[0];
    int q=1,temp;
    for(int i=0;i<k;i++)
    {
        while(temp=Cross(p[i+1]-p[i],p[q+1]-p[q])>0)  q=(q+1)%k;
        ans=max(Length(tubao[q]-tubao[i]),ans);
        if(!temp) ans=max(ans,Length(tubao[q+1]-tubao[i]));
    }
    printf("%d
",(int)(ans*ans+0.5));
}
*/
Point p[1005];
Vector v[105];
int n,f[505][505],vis[505];
;
void init()
{
   for(int i=0;i<n;i++)
      {
          p[i]=p[i]-v[i]*1e-5;
          p[i+n]=p[i+n]+v[i]*1e-5;
      }
   memset(f,0,sizeof(f));
   memset(vis,0,sizeof(vis));
   p[2*n]=Point(0,0);p[2*n+1]=Point(100,100);
   cout<<":7"<<endl;
   for(int i=0;i<=2*n+1;i++)
   for(int j=i+1;j<=2*n+1;j++)
   {
       int flag=1;
       cout<<":9"<<endl;
       for(int k=0;k<n;j++)
          if(SegmentProperIntersection(p[i],p[j],p[k],p[k+n]))
               {cout<<13<<endl;flag=0;break;}
       cout<<":10"<<endl;
       if(flag) f[i][j]=f[j][i]=1;
       cout<<":8"<<endl;
   }
   cout<<":9"<<endl;
}

int dfs(int cur)
{
    if(!vis[cur]) vis[cur]=1;
    else return 0;
    for(int i=0;i<=2*n+1;i++)
       if(!vis[i]&&f[cur][i]&&i!=cur)
            {
                if(i==2*n+1) return 1;
                else if(dfs(i)) return 1;
                cout<<":3"<<endl;
            }
    cout<<":4"<<endl;
    return 0;
}

void solve()
{
    if(dfs(2*n)) printf("yes
");
    else printf("no
");
}

int main()
{
    while(~scanf("%d",&n)&&n)
    {
       for(int i=0;i<n;i++)
            {
                scanf("%d %d",&p[i].x,&p[i].y);
                scanf("%d %d",&p[i+n].x,&p[i+n].y);
                v[i]=(p[i+n]-p[i])/(Length(p[i+n]-p[i]));
            }
       cout<<":1"<<endl;
       init();
       cout<<":2"<<endl;
       solve();
    }
    return 0;
}