Video Surveillance POJ - 1474 （半平面交）
Video Surveillance
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)\n",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\n",++cases); 85 if (halfplaneinsert()) printf("Surveillance is possible.\n"); 86 else printf("Surveillance is impossible.\n"); 87 printf("\n"); 88 } 89 }
