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。

  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 }

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