Video Surveillance POJ - 1474 （半平面交）

Posted 2021-03-11 vampire6

Video Surveillance

 POJ - 1474 

题意：多边形是否有内核

思路：平面半交题，注意直线的存入顺序

 1 // 
 2 // Created by HJYL on 2020/2/6.
 3 //
 4 #include<iostream>
 5 #include<stdio.h>
 6 #include<algorithm>
 7 #include<math.h>
 8 using namespace std;
 9 typedef long long ll;
10 typedef double db;
11 const int N=207;
12 const db eps=1e-7;
13 int sign(db k){if(k>eps) return 1;if(k<-eps) return -1; return 0;}
14 int dcmp(db k1,db k2){return sign(k1-k2);}
15 struct point{
16     db x,y;
17     point operator - (const point &k)const{
18         return (point){x-k.x,y-k.y};
19     }
20     point operator + (const point &k)const{
21         return (point){x+k.x,y+k.y};
22     }
23     point operator * (const db &k)const{
24         return (point){x*k,y*k};
25     }
26     point operator / (const db &k)const{
27         return (point){x/k,y/k};
28     }
29     db angle(){return atan2(y,x);}
30     void print(){printf("(%f,%f)
",x,y);}
31 }P[N];
32 db cross(point k1,point k2){
33     return k1.x*k2.y-k1.y*k2.x;
34 }
35 db dot(point k1,point k2){
36     return k1.x*k2.x+k1.y*k2.y;
37 }
38 struct line{
39     point s,e;
40     db angle(){return (e-s).angle();}
41 }L[N],dq[N];
42 point getLL(line k1,line k2){
43     db w1=cross(k1.s-k2.s,k2.e-k2.s),w2=cross(k2.e-k2.s,k1.e-k2.s);
44     return (k1.s*w2+k1.e*w1)/(w1+w2);
45 }
46 bool onRight(line k,point p){
47     return sign((cross(p-k.s,k.e-k.s)))>0;
48 }
49 bool cmp(line k1,line k2){
50     if(dcmp(k1.angle(),k2.angle())==0) return onRight(k2,k1.s);
51     return k1.angle()<k2.angle();
52 }
53 int tot;
54 bool halfplaneinsert(){
55     sort(L+1,L+1+tot,cmp);
56     int cnt=0;
57     for(int i=1;i<=tot;i++){
58         if(i<tot&&dcmp(L[i].angle(),L[i+1].angle())==0) continue;
59         L[++cnt]=L[i];
60     }
61     int tail=-1,head=0;
62     for(int i=1;i<=cnt;i++){
63         while(tail-head>=1&&onRight(L[i],getLL(dq[tail],dq[tail-1]))) tail--;
64         while(tail-head>=1&&onRight(L[i],getLL(dq[head],dq[head+1]))) head++;
65         dq[++tail]=L[i];
66     }
67     while(tail-head>=1&&onRight(dq[head],getLL(dq[tail],dq[tail-1]))) tail--;
68     while(tail-head>=1&&onRight(dq[tail],getLL(dq[head],dq[head+1]))) head++;
69     return tail-head>=2;
70 }
71 int main()
72 {
73     int cases=0;
74     int n;
75     while(~scanf("%d",&n)&&n) {
76             for (int i = 1; i <= n; i++) {
77                 scanf("%lf%lf", &P[i].x, &P[i].y);
78             }
79             tot = 0;
80             for (int i = n; i > 1; i--) {
81                 L[++tot] = (line) {P[i], P[i - 1]};
82             }
83             L[++tot] = (line) {P[1], P[n]};
84             printf("Floor #%d
",++cases);
85             if (halfplaneinsert()) printf("Surveillance is possible.
");
86             else printf("Surveillance is impossible.
");
87             printf("
");
88         }
89 }