LA 2797
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\n",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\n");
else printf("yes\n");
}
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\n",(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\n");
else printf("no\n");
}
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;
}
你说，我们都会幸福的，对吧？
