求一个多边形是否完全在另一个凸多边形内。
乍一看，好像要判点在多边形内，但复杂度不允许，仔细一想，可以把两个多边形的点混起来求一个共同的凸包，如果共同的凸包依旧是原来凸包上的点，说明是。
1 #include <iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #include<stdlib.h> 6 #include<vector> 7 #include<cmath> 8 #include<queue> 9 #include<set> 10 using namespace std; 11 #define N 200010 12 #define LL long long 13 #define INF 0xfffffff 14 const double eps = 1e-8; 15 const double pi = acos(-1.0); 16 const double inf = ~0u>>2; 17 struct point 18 { 19 double x,y; 20 int flag; 21 point(double x=0,double y=0):x(x),y(y){} 22 }p[N],ch[N],q[N]; 23 typedef point pointt; 24 point operator -(point a,point b) 25 { 26 return point(a.x-b.x,a.y-b.y); 27 } 28 int dcmp(double x) 29 { 30 if(fabs(x)<eps) return 0; 31 return x<0?-1:1; 32 } 33 double cross(point a,point b) 34 { 35 return a.x*b.y-a.y*b.x; 36 } 37 double mul(point a,point b,point c) 38 { 39 return cross(b-a,c-a); 40 } 41 double dis(point a) 42 { 43 return sqrt(a.x*a.x+a.y*a.y); 44 } 45 bool cmp(point a,point b) 46 { 47 if(mul(p[0],a,b)==0) 48 return dis(p[0]-a)<dis(p[0]-b); 49 return mul(p[0],a,b)>0; 50 } 51 int Graham(int n) 52 { 53 int i,k = 0,top; 54 point tmp; 55 for(i = 0 ; i < n; i++) 56 { 57 if(p[i].y<p[k].y||(p[i].y==p[k].y&&p[i].x<p[k].x)) 58 k = i; 59 } 60 if(k!=0) 61 { 62 tmp = p[0]; 63 p[0] = p[k]; 64 p[k] = tmp; 65 } 66 sort(p+1,p+n,cmp); 67 ch[0] = p[0]; 68 ch[1] = p[1]; 69 top = 1; 70 for(i = 2; i < n ; i++) 71 { 72 while(top>0&&dcmp(mul(ch[top-1],ch[top],p[i]))<0) 73 top--; 74 top++; 75 ch[top] = p[i]; 76 } 77 return top; 78 } 79 int dot_online_in(point p,point l1,point l2) 80 { 81 return !dcmp(mul(p,l1,l2))&&(l1.x-p.x)*(l2.x-p.x)<eps&&(l1.y-p.y)*(l2.y-p.y)<eps; 82 } 83 int main() 84 { 85 int n,m,i; 86 cin>>n; 87 for(i =0 ; i < n; i++) 88 { 89 scanf("%lf%lf",&p[i].x,&p[i].y); 90 p[i].flag = 0; 91 } 92 cin>>m; 93 for(i = n ; i < n+m; i++) 94 { 95 scanf("%lf%lf",&p[i].x,&p[i].y); 96 p[i].flag = 1; 97 q[i-n] = p[i]; 98 } 99 int tn = Graham(n+m); 100 ch[tn+1] = ch[0]; 101 int ff = 1; 102 for(i = 0 ; i < m ; i++) 103 { 104 if(dot_online_in(q[i],ch[tn],ch[tn+1])) 105 { 106 ff = 0; 107 //cout<<p[i].x<<" "<<p[i].y<<endl; 108 break; 109 } 110 } 111 if(!ff) 112 { 113 puts("NO"); 114 return 0; 115 } 116 for(i = 0; i <= tn ; i++) 117 if(ch[i].flag) {ff = 0;break;} 118 if(ff) puts("YES"); 119 else puts("NO"); 120 return 0; 121 }