推薦部落格 ：借鑑
分別列舉每個凸包的每條邊，再列舉另乙個凸包的點，求得一對對踵點，與當前記錄的值相取捨。
遍歷到下一條邊，從上次便利到的點開始尋找下一對踵點，再取捨......。直至取捨出最長、最短距離。
*/#include
#include
#include
using namespace std;
const int maxn=50001;
struct node
st[maxn],num[maxn];
int n,k,e;
double dis(node p1,node p2)
int multi(node p1,node p2,node p3)//叉積
bool cmp(node p1,node p2)
}double rotating_calipers()
return ans;
}int main()
swap(num[0],num[k]);
//printf("%d\n",k);
graham();
printf("%.0lf\n",rotating_calipers());
return 0;
}
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