Pick-up sticks

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 15265		Accepted: 5752

Description

Stan has n sticks of various length. He throws them one at a time on the floor in a random way. After finishing throwing, Stan tries to find the top sticks, that is these sticks such that there is no stick on top of them. Stan has noticed that the last thrown stick is always on top but he wants to know all the sticks that are on top. Stan sticks are very, very thin such that their thickness can be neglected.

Input

Input consists of a number of cases. The data for each case start with 1 <= n <= 100000, the number of sticks for this case. The following n lines contain four numbers each, these numbers are the planar coordinates of the endpoints of one stick. The sticks are listed in the order in which Stan has thrown them. You may assume that there are no more than 1000 top sticks. The input is ended by the case with n=0. This case should not be processed.

Output

For each input case, print one line of output listing the top sticks in the format given in the sample. The top sticks should be listed in order in which they were thrown.

The picture to the right below illustrates the first case from input.

Sample Input

Sample Output

Top sticks: 2, 4, 5.
Top sticks: 1, 2, 3.

Hint

Huge input,scanf is recommended.

这题判断是否相交就行

把bin巨的模板打了一遍。。。

  1 #include<cstdio>
  2 #include<iostream>
  3 #include<cstring>
  4 #include<cmath>
  5 using namespace std;
  6 const double eps=1e-8;
  7 const double INF=1e20;
  8 const double PI=acos(-1.0);
  9 const int maxp=1010;
 10 int sgn(double x){
 11     if(fabs(x)<eps) return 0;
 12     if(x<0) return -1;
 13     else return 1;
 14 }
 15 inline double sqr(double x){return x*x;}
 16 struct Point{
 17     double x,y;
 18     Point(){}
 19     Point(double _x,double _y){
 20         x=_x;
 21         y=_y;
 22     }
 23     void input(){
 24         scanf("%lf %lf",&x,&y);
 25     }
 26     void output(){
 27         printf("%.2f %.2f
",x,y);
 28     }
 29     bool operator == (const Point &b)const{
 30         return sgn(x-b.x) == 0 && sgn(y-b.y)== 0;
 31     }
 32     bool operator < (const Point &b)const{
 33         return sgn(x-b.x)==0?sgn(y-b.y)<0:x<b.x;
 34     }
 35     Point operator - (const Point &b)const{
 36         return Point(x-b.x,y-b.y);
 37     }
 38     //叉积
 39     double operator ^ (const Point &b)const{
 40         return x*b.y-y*b.x;
 41     }
 42     //点积
 43     double operator * (const Point &b)const{
 44         return x*b.x+y*b.y;
 45     }
 46     //返回长度
 47     double len(){
 48         return hypot(x,y);
 49     }
 50     //返回长度的平方
 51     double len2(){
 52         return x*x+y*y;
 53     }
 54     //返回两点的距离
 55     double distance(Point p){
 56         return hypot(x-p.x,y-p.y);
 57     }
 58     Point operator + (const Point &b)const{
 59         return Point(x+b.x,y+b.y);
 60     }
 61     Point operator * (const double &k)const{
 62         return Point(x*k,y*k);
 63     }
 64     Point operator / (const double &k)const{
 65         return Point(x/k,y/k);
 66     }
 67 
 68     //计算pa和pb的夹角
 69     //就是求这个点看a,b所成的夹角
 70     ///LightOJ1202
 71     double rad(Point a,Point b){
 72         Point p=*this;
 73         return fabs(atan2(fabs((a-p)^(b-p)),(a-p)*(b-p)));
 74     }
 75     //化为长度为r的向量
 76     Point trunc(double r){
 77         double l=len();
 78         if(!sgn(l)) return *this;
 79         r/=l;
 80         return Point(x*r,y*r);
 81     }
 82     //逆时针转90度
 83     Point rotleft(){
 84         return Point(-y,x);
 85     }
 86     //顺时针转90度
 87     Point rotright(){
 88         return Point(y,-x);
 89     }
 90     //绕着p点逆时针旋转angle
 91     Point rotate(Point p,double angle){
 92         Point v=(*this) -p;
 93         double c=cos(angle),s=sin(angle);
 94         return Point(p.x+v.x*c-v.y*s,p.y+v.x*s+v.y*c);
 95     }
 96 };
 97 
 98 struct Line{
 99     Point s,e;
100     Line(){}
101     Line(Point _s,Point _e){
102         s=_s;
103         e=_e;
104     }
105     bool operator==(Line v){
106         return (s==v.s)&&(e==v.e);
107     }
108     //根据一个点和倾斜角angle确定直线，0<=angle<pi
109     Line(Point p,double angle){
110         s=p;
111         if(sgn(angle-PI/2)==0){
112             e=(s+Point(0,1));
113         }
114         else{
115             e=(s+Point(1,tan(angle)));
116         }
117     }
118     //ax+by+c=0;
119     Line(double a,double b,double c){
120         if(sgn(a)==0){
121             s=Point(0,-c/b);
122             e=Point(1,-c/b);
123         }
124         else if(sgn(b)==0){
125             s=Point(-c/a,0);
126             e=Point(-c/a,1);
127         }
128         else{
129             s=Point(0,-c/b);
130             e=Point(1,(-c-a)/b);
131         }
132     }
133     void input(){
134         s.input();
135         e.input();
136     }
137     void adjust(){
138         if(e<s) swap(s,e);
139     }
140     //求线段长度
141     double length(){
142         return s.distance(e);
143     }
144     //返回直线倾斜角 0<=angle<pi
145     double angle(){
146         double k=atan2(e.y-s.y,e.x-s.x);
147         if(sgn(k)<0) k+=PI;
148         if(sgn(k-PI)==0) k-=PI;
149         return k;
150     }
151     //点和直线的关系
152     //1 在左侧
153     //2 在右侧
154     //3 在直线上
155     int relation(Point p){
156         int c=sgn((p-s)^(e-s));
157         if(c<0) return 1;
158         else if(c>0) return 2;
159         else return 3;
160     }
161     //点在线段上的判断
162     bool pointonseg(Point p){
163         return sgn((p-s)^(e-s))==0&&sgn((p-s)*(p-e))<=0;
164     }
165     //两向量平行(对应直线平行或重合)
166     bool parallel(Line v){
167         return sgn((e-s)^(v.e-v.s))==0;
168     }
169     //两线段相交判断
170     //2 规范相交
171     //1 非规范相交
172     //0 不相交
173     int segcrossseg(Line v){
174         int d1=sgn((e-s)^(v.s-s));
175         int d2=sgn((e-s)^(v.e-s));
176         int d3=sgn((v.e-v.s)^(s-v.s));
177         int d4=sgn((v.e-v.s)^(e-v.s));
178         if((d1^d2)==-2&&(d3^d4)==-2) return 2;
179         return (d1==0&&sgn((v.s-s)*(v.s-e))<=0||
180                 d2==0&&sgn((v.e-s)*(v.e-e))<=0||
181                 d3==0&&sgn((s-v.s)*(s-v.e))<=0||
182                 d4==0&&sgn((e-v.s)*(e-v.e))<=0);
183     }
184     //直线和线段相交判断
185     //-*this line -v seg
186     //2 规范相交
187     //1 非规范相交
188     //0 不相交
189     int linecrossseg(Line v){
190         int d1=sgn((e-s)^(v.s-s));
191         int d2=sgn((e-s)^(v.e-s));
192         if((d1^d2)==-2) return 2;
193         return (d1==0||d2==0);
194     }
195     //两直线关系
196     //0 平行
197     //1 重合
198     //2 相交
199     int linecrossline(Line v){
200         if((*this).parallel(v))
201             return v.relation(s)==3;
202         return 2;
203     }
204     //求两直线的交点
205     //要保证两直线不平行或重合
206     Point crosspoint(Line v){
207         double a1=(v.e-v.s)^(s-v.s);
208         double a2=(v.e-v.s)^(e-v.s);
209         return Point((s.x*a2-e.x*a1)/(a2-a1),(s.y*a2-e.y*a1)/(a2-a1));
210     }
211     //点到直线的距离
212     double dispointtoline(Point p){
213         return fabs((p-s)^(e-s))/length();
214     }
215     //点到线段的距离
216     double dispointtoseg(Point p){
217         if(sgn((p-s)*(e-s))<0||sgn((p-e)*(s-e))<0)
218             return min(p.distance(s),p.distance(e));
219         return dispointtoline(p);
220     }
221     //返回线段到线段的距离
222     //前提是两线段不相交，相交距离就是0了
223     double dissegtoseg(Line v){
224         return min(min(dispointtoseg(v.s),dispointtoseg(v.e)),min(v.dispointtoseg(s),v.dispointtoseg(e)));
225     }
226     //返回点P在直线上的投影
227     Point lineprog(Point p){
228         return s+(((e-s)*((e-s)*(p-s)))/((e-s).len2()));
229     }
230     //返回点P关于直线的对称点
231     Point symmetrypoint(Point p){
232         Point q=lineprog(p);
233         return Point(2*q.x-p.x,2*q.y-p.y);
234     }
235 };
236 
237 Line L[100005];
238 int book[100005];
239 int n;
240 
241 bool Check(Line a,Line b){
242     if(((a.s-a.e)^(b.s-a.e))*((a.s-a.e)^(b.e-a.e))>0) return false;
243     if(((b.s-b.e)^(a.s-b.e))*((b.s-b.e)^(a.e-b.e))>0) return false;
244     return true;
245 }
246 
247 
248 int main(){
249     while(~scanf("%d",&n)){
250         if(!n) break;
251         memset(book,0,sizeof(book));
252         for(int i=1;i<=n;i++){
253             scanf("%lf %lf %lf %lf",&L[i].s.x,&L[i].s.y,&L[i].e.x,&L[i].e.y);
254         }
255         int flag;
256         for(int i=1;i<=n;i++){
257             flag=0;
258             for(int j=i+1;j<=n;j++){
259                 if(Check(L[i],L[j])){
260                     flag=1;
261                     break;
262                 }
263             }
264             if(flag){
265                 book[i]=1;
266             }
267         }
268         flag=0;
269         printf("Top sticks: ");
270         for(int i=1;i<=n;i++){
271             if(!book[i]){
272                 if(!flag) flag=1;
273                 else printf(", ");
274                 printf("%d",i);
275 
276             }
277         }
278         printf(".
");
279     }
280     return 0;
281 }

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