This article is made by Jason-Cow.

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利用一下叉积和栈

1 int Andrew(D*R,int n,D*A){ 2     int m=0; 3     sort(R+1,R+n+1);//for(int i=1;i<=n;i++)cout<
     
      <<" "<
      
       <
       
        =2 && Cross(A[m]-A[m-1],R[i]-A[m-1])<=0)m--; 6         A[++m]=R[i]; 7     } 8     int k=m; 9     for(int i=n-1;i>=1;i--){10         while(m>k  && Cross(A[m]-A[m-1],R[i]-A[m-1])<=0)m--;11         A[++m]=R[i];12     }13     return n>1?m-1:m;14 }