题意:给你一些点,求最远点对
思路:先求凸包,然后旋转卡壳模板,不断枚举边,找凸包上的最远点,更新答案
感谢:http://hzwer.com/4224.html
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<cstdlib> 5 #include<set> 6 #include<ctime> 7 #include<vector> 8 #include<queue> 9 #include<algorithm> 10 #include<map> 11 #include<cmath> 12 #define eps 1e-8 13 #define inf 1000000000 14 #define pa pair<int,int> 15 #define ll long long 16 using namespace std; 17 int read() 18 { 19 int x=0,f=1;char ch=getchar(); 20 while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} 21 while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} 22 return x*f; 23 } 24 int n,top; 25 double ans; 26 double sqr(double x) 27 { 28 29 return x*x; 30 } 31 struct P{ 32 double x,y; 33 P(){} 34 P(double _x,double _y):x(_x),y(_y){} 35 friend P operator +(P a,P b){ 36 return P(a.x+b.x,a.y+b.y); 37 } 38 friend P operator -(P a,P b){ 39 return P(a.x-b.x,a.y-b.y); 40 } 41 friend double operator*(P a,P b){ 42 return a.x*b.y-a.y*b.x; 43 } 44 friend double operator/(P a,P b){ 45 return a.x*b.x+a.y*b.y; 46 } 47 friend bool operator==(P a,P b){ 48 return fabs(a.x-b.x)<eps&&fabs(a.y-b.y)<eps; 49 } 50 friend bool operator!=(P a,P b){ 51 return !(a==b); 52 } 53 friend bool operator<(P a,P b){ 54 if(fabs(a.y-b.y)<eps)return a.x<b.x; 55 return a.y<b.y; 56 } 57 friend double dis2(P a){ 58 return sqr(a.x)+sqr(a.y); 59 } 60 friend void print(P a){ 61 printf("%.2lf %.2lf ",a.x,a.y); 62 } 63 }p[50005],q[50005]; 64 bool cmp(P a,P b) 65 { 66 if(fabs((b-p[1])*(a-p[1]))<eps)return dis2(a-p[1])<dis2(b-p[1]); 67 return (a-p[1])*(b-p[1])>0; 68 } 69 void graham()//求求凸包 原来点用p存 求完凸包放在q里面 70 { 71 for(int i=1;i<=n;i++) 72 if(p[i]<p[1])swap(p[i],p[1]); 73 sort(p+2,p+n+1,cmp); 74 q[++top]=p[1];q[++top]=p[2]; 75 for(int i=3;i<=n;i++) 76 { 77 while((q[top]-q[top-1])*(p[i]-q[top-1])<eps&&top>1)top--; 78 q[++top]=p[i]; 79 } 80 } 81 void RC() 82 { 83 q[top+1]=q[1]; 84 int now=2; 85 for(int i=1;i<=top;i++)//枚举边 86 { 87 while((q[i+1]-q[i])*(q[now]-q[i])<(q[i+1]-q[i])*(q[now+1]-q[i])) 88 { 89 now++; 90 if(now==top+1)now=1; 91 }//q[now]为凸包上距离当前边最远的点 92 ans=max(ans,dis2(q[now]-q[i])); 93 } 94 } 95 int main() 96 { 97 n=read(); 98 for(int i=1;i<=n;i++) 99 p[i].x=read(),p[i].y=read(); 100 graham(); 101 RC(); 102 printf("%d",(int)ans); 103 return 0; 104 }