CodeForces 166B （凸包）

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。

 #include <iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<stdlib.h>
 #include<vector>
 #include<cmath>
 #include<queue>
 #include<set>
 using namespace std;
 #define N 200010
 #define LL long long
 #define INF 0xfffffff
 const double eps = 1e-;
 const double pi = acos(-1.0);
 const double inf = ~0u>>;
 struct point
 {
     double x,y;
     int flag;
     point(double x=,double y=):x(x),y(y){}
 }p[N],ch[N],q[N];
 typedef point pointt;
 point operator -(point a,point b)
 {
     return point(a.x-b.x,a.y-b.y);
 }
 int dcmp(double x)
 {
     if(fabs(x)<eps) return ;
     return x<?-:;
 }
 double cross(point a,point b)
 {
     return a.x*b.y-a.y*b.x;
 }
 double mul(point a,point b,point c)
 {
     return cross(b-a,c-a);
 }
 double dis(point a)
 {
     return sqrt(a.x*a.x+a.y*a.y);
 }
 bool cmp(point a,point b)
 {
     if(mul(p[],a,b)==)
     return dis(p[]-a)<dis(p[]-b);
     return mul(p[],a,b)>;
 }
 int Graham(int n)
 {
     int i,k = ,top;
     point tmp;
     for(i =  ; i < n; i++)
     {
         if(p[i].y<p[k].y||(p[i].y==p[k].y&&p[i].x<p[k].x))
             k = i;
     }
     if(k!=)
     {
         tmp = p[];
         p[] = p[k];
         p[k] = tmp;
     }
     sort(p+,p+n,cmp);
     ch[] = p[];
     ch[] = p[];
     top = ;
     for(i = ; i < n ; i++)
     {
         while(top>&&dcmp(mul(ch[top-],ch[top],p[i]))<)
             top--;
         top++;
         ch[top] = p[i];
     }
     return top;
 }
 int dot_online_in(point p,point l1,point l2)
 {
     return !dcmp(mul(p,l1,l2))&&(l1.x-p.x)*(l2.x-p.x)<eps&&(l1.y-p.y)*(l2.y-p.y)<eps;
 }
 int main()
 {
     int n,m,i;
     cin>>n;
     for(i = ; i < n; i++)
     {
         scanf("%lf%lf",&p[i].x,&p[i].y);
         p[i].flag = ;
     }
     cin>>m;
     for(i = n ; i < n+m; i++)
     {
         scanf("%lf%lf",&p[i].x,&p[i].y);
         p[i].flag = ;
         q[i-n] = p[i];
     }
     int tn = Graham(n+m);
     ch[tn+] = ch[];
     int ff = ;
     for(i =  ; i < m ; i++)
     {
         if(dot_online_in(q[i],ch[tn],ch[tn+]))
         {
             ff  = ;
             //cout<<p[i].x<<" "<<p[i].y<<endl;
             break;
         }
     }
     if(!ff)
     {
         puts("NO");
         return ;
     }
     for(i = ; i <= tn ; i++)
     if(ch[i].flag) {ff = ;break;}
     if(ff) puts("YES");
     else puts("NO");
     return ;
 }